Selina Concise Solutions for ICSE Class 10 Chemistry Chapter 5 Mole Concept and Stoichiometry

Exercise 5(A)

Solution 1.
Answer: (a) Gay-Lussac's law states that when gases react, they do so in volumes which bear a simple ratio to one another, and to the volume of the gaseous product, provided that all the volumes are measured at the same temperature and pressure.
(b) Avogadro's law states that equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules.
In simple words: These are two important gas laws that help us understand how gases behave and combine with each other in fixed ratios.

📝 Teacher's Note: Use everyday examples like mixing different gases in balloons to demonstrate these laws. Students often confuse these two laws, so emphasize that Gay-Lussac's is about reacting volumes while Avogadro's is about equal volumes containing equal molecules.

🎯 Exam Tip: Always mention "same temperature and pressure" when stating these laws - it's a key point examiners look for in definitions.

Solution 2.
Answer: a) The number of atoms in a molecule of an element is called its atomicity. Atomicity of Hydrogen is 2, phosphorus is 4 and sulphur is 8.
b) \( N_2 \) means 1 molecule of nitrogen and 2N means two atoms of nitrogen. \( N_2 \) can exist independently but 2N cannot exist independently.
In simple words: Atomicity tells us how many atoms stick together to form one molecule. Free atoms like 2N cannot exist alone in nature - they always need to be paired up.

📝 Teacher's Note: Use physical models or drawings to show the difference between molecules and individual atoms. Emphasize that most non-metal elements exist as molecules, not single atoms.

🎯 Exam Tip: Remember the key atomicities: H₂=2, O₂=2, P₄=4, S₈=8. Always explain why free atoms cannot exist independently.

Solution 3.
Answer: (a) This is due to Avogadro's Law which states Equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules. Now volume of hydrogen gas = volume of helium gas, so n molecules of hydrogen = n molecules of helium gas, \( nH_2 = nHe \). 1 mol. of hydrogen has 2 atoms of hydrogen and 1 molecule of helium has 1 atom of helium. Therefore 2H = He. Therefore atoms in hydrogen is double the atoms of helium.
(b) For a given volume of gas under given temperature and pressure, a change in any one of the variable i.e., pressure or temperature changes the volume.
(c) Inflating a balloon seems violating Boyle's law as volume is increasing with increase in pressure. Since the mass of gas is also increasing.
In simple words: Equal volumes of different gases have the same number of molecules, but hydrogen molecules contain 2 atoms each while helium has only 1 atom per molecule.

📝 Teacher's Note: Draw molecular diagrams showing H₂ molecules vs He atoms to visualize why hydrogen has double the atoms. For part (c), explain that Boyle's law assumes constant amount of gas, but balloon inflation adds more gas.

🎯 Exam Tip: Always distinguish between molecules and atoms in your explanations. For gas law violations, identify which variable is being kept constant and which is changing.

Solution 4.
Answer: \( 2H_2 + O_2 \rightarrow 2H_2O \)
2V 1V 2V
From the equation, 2V of hydrogen reacts with 1V of oxygen, so 200cm³ of Hydrogen reacts with = 200/2 = 100 cm³. Hence, the unreacted oxygen is 150 - 100 = 50cm³ of oxygen.
In simple words: Using the balanced equation, we can calculate how much oxygen is needed and how much will be left over after the reaction.

📝 Teacher's Note: Start with the balanced equation and use stoichiometric ratios. Remind students that the limiting reagent determines how much product forms and how much excess reagent remains.

🎯 Exam Tip: Always write the balanced equation first, then use volume ratios from coefficients. Show your calculation steps clearly for full marks.

Solution 5.
Answer: \( CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \)
1V 2V 1V
From equation, 1V of \( CH_4 \) reacts with = 2V of \( O_2 \)
so, 80 cm³ \( CH_4 \) reacts with = 80 × 2 = 160cm³ \( O_2 \)
Remaining \( O_2 \) is 200-160 = 40cm³
From equation, 1V of methane gives 1V of \( CO_2 \)
So, 80 cm³ gives 80cm³ \( CO_2 \) and \( H_2O \) is negligible.
In simple words: Methane burns in oxygen to produce carbon dioxide and water, following a fixed volume ratio from the balanced equation.

📝 Teacher's Note: Explain why water vapor is negligible at room temperature (it condenses). Use this problem to reinforce stoichiometric calculations with gases.

🎯 Exam Tip: State which reagent is limiting and calculate both products formed and excess reagents remaining.

Solution 6.
Answer: \( 2C_2H_2 + 5O_2 \rightarrow 4CO_2 + 2H_2O \) (l)
2V 5V 4V
From equation, 2V of \( C_2H_2 \) requires = 5V of \( O_2 \)
So, for 400ml \( C_2H_2 \), \( O_2 \) required = 400 × 5/2 = 1000 ml
Similarly, 2V of \( C_2H_2 \) gives = 4V of \( CO_2 \)
So, 400ml of \( C_2H_2 \) gives \( CO_2 \) = 400 × 4/2 = 800ml
In simple words: Acetylene burns completely in oxygen to produce carbon dioxide and water in fixed volume ratios.

📝 Teacher's Note: Emphasize the importance of balancing equations correctly before applying stoichiometry. Point out that water is liquid (l) at room conditions.

🎯 Exam Tip: Use cross-multiplication method for ratio calculations. Always check if your answer makes chemical sense.

Solution 7.
Answer: Balanced chemical equation: \( H_2S_{(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)} + S_{(s)} \)
1mole 1mole 2moles 1mol
112cm³ 120cm³
(i) At STP, 1 mole gas occupies 22.4 L. As 1 mole \( H_2S \) gas produces 2 moles HCl gas, 22.4 L \( H_2S \) gas produces 22.4 × 2 = 44.8 L HCl gas. Hence, 112 cm³ \( H_2S \) gas will produce 112 × 2 = 224 cm³ HCl gas.
(ii) 1 mole \( H_2S \) gas consumes 1 mole \( Cl_2 \) gas. This means 22.4 L \( H_2S \) gas consumes 22.4 L \( Cl_2 \) gas at STP. Hence, 112 cm³ \( H_2S \) gas consumes 112 cm³ \( Cl_2 \) gas. 120 cm³ - 112 cm³ = 8 cm³ \( Cl_2 \) gas remains unreacted. Thus, the composition of the resulting mixture is 224 cm³ HCl gas + 8 cm³ \( Cl_2 \) gas.
In simple words: When hydrogen sulfide reacts with chlorine, it produces hydrogen chloride gas and solid sulfur, with some chlorine left over.

📝 Teacher's Note: Emphasize that sulfur is produced as a solid and doesn't contribute to gas volume. Teach students to identify limiting and excess reagents systematically.

🎯 Exam Tip: Always identify which gas is limiting, calculate products based on the limiting reagent, then find excess reagent remaining.

Solution 8.
Answer: \( 2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O \)
2V 7V 4V
Now from equation, 2V of ethane reacts with = 7V of oxygen
So, 600cc of ethane reacts with = 600 × 7/2 = 2100cc
Hence, unused \( O_2 \) is = 2500 - 2100 = 400 cc
From 2V of ethane = 4V of \( CO_2 \) is produced
So, 600cc of ethane will produce = 4 × 600/2 = 1200cc \( CO_2 \)
In simple words: Ethane burns in oxygen to produce carbon dioxide and water, with some oxygen remaining unused.

📝 Teacher's Note: Point out the 7:2 ratio of oxygen to ethane - this is higher than methane because ethane has more carbon and hydrogen atoms to burn.

🎯 Exam Tip: Double-check your balanced equation coefficients - ethane combustion requires 7 oxygen molecules for every 2 ethane molecules.

Solution 9.
Answer: \( C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O \)
1V 3V
1 litre 3 litre
\( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \)
\( V_2 = \frac{P_1V_1T_2}{P_2T_1} = \frac{380 \times 33 \times 273}{549 \times 760} = 8.25 \) litres
In simple words: Using gas laws, we can calculate how the volume of ethylene changes when temperature and pressure conditions change.

📝 Teacher's Note: This combines stoichiometry with gas law calculations. Make sure students convert pressure units consistently (mmHg to mmHg or atm to atm).

🎯 Exam Tip: Always convert temperature to Kelvin and ensure pressure units match. Use the combined gas law formula correctly.

Solution 10.
Answer: \( CH_4 + 2Cl_2 \rightarrow CH_2Cl_2 + 2HCl \)
1V 2V 1V 2V
From equation, 1V of \( CH_4 \) gives = 2V HCl
so, 40 ml of methane gives = 80 ml HCl
For 1V of methane = 2V of \( Cl_2 \) required
So, for 40ml of methane = 40 × 2 = 80 ml of \( Cl_2 \)
In simple words: Methane reacts with chlorine to form dichloromethane and hydrogen chloride gas in fixed volume ratios.

📝 Teacher's Note: Explain that this is a substitution reaction where chlorine replaces hydrogen atoms. The product CH₂Cl₂ is dichloromethane, a common solvent.

🎯 Exam Tip: Balance the equation carefully - each methane molecule needs two chlorine molecules to form the dichloride product.

Solution 11.
Answer: \( C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O \)
1V 5V 3V
From equation, 5V of \( O_2 \) required = 1V of propane
so, 100 cm³ of \( O_2 \) will require = 20 cm³ of propane
In simple words: Propane burns in oxygen following a 1:5 volume ratio, so we can calculate how much propane burns completely with a given amount of oxygen.

📝 Teacher's Note: Propane is commonly used in gas cylinders and stoves. The 5:1 oxygen to propane ratio shows why complete combustion needs plenty of air.

🎯 Exam Tip: Remember propane's formula C₃H₈ and that it needs 5 oxygen molecules for complete combustion. Work backwards from oxygen to find propane needed.

Solution 12.
Answer: \( 2NO + O_2 \rightarrow 2NO_2 \)
2V 1V 2V
From equation, 1V of \( O_2 \) reacts with = 2V of NO
200cm³ oxygen will react with = 200 × 2 = 400 cm³ NO
Hence, remaining NO is 450 - 400 = 50 cm³
\( NO_2 \) produced = 400cm³ because 1V oxygen gives 2V \( NO_2 \)
Total mixture = 400 + 50 = 450 cm³
In simple words: Nitrogen monoxide reacts with oxygen to form nitrogen dioxide, with some nitrogen monoxide left unreacted.

📝 Teacher's Note: This reaction is important in air pollution chemistry. NO₂ is the brown gas that contributes to smog formation in cities.

🎯 Exam Tip: Identify oxygen as the limiting reagent, calculate products based on oxygen available, then find unreacted NO remaining.

Solution 13.
Answer: \( 2CO + O_2 \rightarrow 2CO_2 \)
2V 1V 2V
2V of CO requires = 1V of \( O_2 \)
so, 100 litres of CO requires = 50 litre of \( O_2 \)
In simple words: Carbon monoxide burns in oxygen to form carbon dioxide, following a 2:1 volume ratio.

📝 Teacher's Note: Emphasize that this reaction converts toxic CO to less harmful CO₂. This is why car catalytic converters are important for reducing emissions.

🎯 Exam Tip: Remember the 2:1 ratio of CO to O₂. This is a simple stoichiometry problem - divide CO volume by 2 to get required oxygen volume.

Solution 14.
Answer: \( 4NH_3 + 5O_2 \rightarrow 4NO + 6H_2O \)
4V 5V 4V
9 litres of reactants gives 4 litres of NO
So, 27 litres of reactants will give = 27 × 4/9 = 12 litres of NO
In simple words: Ammonia reacts with oxygen to produce nitrogen monoxide and water vapor, with the volume of products being less than reactants.

📝 Teacher's Note: This is the Ostwald process for making nitric acid. Point out that water vapor may condense, so the actual gas volume might be even less.

🎯 Exam Tip: Add up total reactant volumes (4+5=9) and compare with product gas volume (4V NO only, since H₂O is often liquid).

Solution 15.
Answer: \( H_2 + Cl_2 \rightarrow 2HCl \)
1V 1V 2V
Since 1V hydrogen requires 1V of chlorine and 4cm³ of \( H_2 \) remained behind so the mixture had 16 cm³ hydrogen and 16 cm³ chlorine.
Therefore Resulting mixture is \( H_2 \) = 4cm³, HCl = 32cm³
In simple words: Hydrogen and chlorine react in equal volumes to form hydrogen chloride gas, with some hydrogen left over.

📝 Teacher's Note: Work backwards from the unreacted hydrogen to find the original composition. This teaches students to analyze reaction completion.

🎯 Exam Tip: If 4 cm³ H₂ remains unreacted, then 16 cm³ H₂ must have reacted (making 32 cm³ HCl). Original mixture had 20 cm³ each of H₂ and Cl₂.

Solution 16.
Answer: \( CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \)
1V 2V 1V
\( 2C_2H_2 + 5O_2 \rightarrow 4CO_2 + 2H_2O \)
2V 5V 4V
From the equations, we can see that 1V \( CH_4 \) requires oxygen = 2V \( O_2 \). So, 10cm³ \( CH_4 \) will require = 20 cm³ \( O_2 \). Similarly 2V \( C_2H_2 \) requires = 5V \( O_2 \). So, 10 cm³ \( C_2H_2 \) will require = 25 cm³ \( O_2 \). Now, 20V \( O_2 \) will be present in 100V air and 25V \( O_2 \) will be present in 125V air, so the volume of air required is 225cm³
In simple words: Both gases need oxygen for combustion, and since air is only 20% oxygen, we need 5 times the oxygen volume as air volume.

📝 Teacher's Note: Remind students that air is approximately 20% oxygen, so they need to multiply oxygen requirement by 5 to get air volume needed.

🎯 Exam Tip: Calculate oxygen needed for each gas separately, add them up, then multiply by 5 to convert oxygen volume to air volume.

Solution 17.
Answer: \( C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O \)
\( 2C_4H_{10} + 13O_2 \rightarrow 8CO_2 + 10H_2O \)
60 ml of propane (\( C_3H_8 \)) gives 3 × 60 = 180 ml \( CO_2 \)
40 ml of butane (\( C_4H_{10} \)) gives = 8 × 40/2 = 160 ml of \( CO_2 \)
Total carbon dioxide produced = 340 ml
So, when 10 litres of the mixture is burnt = 34 litres of \( CO_2 \) is produced.
In simple words: Both propane and butane produce carbon dioxide when burned, and we can calculate the total CO₂ from the mixture composition.

📝 Teacher's Note: This problem involves mixture stoichiometry. Teach students to handle each component separately then combine results.

🎯 Exam Tip: Calculate CO₂ from each gas separately using their individual balanced equations, then add the results for total CO₂ production.

Solution 18.
Answer: \( 2C_2H_2(g) + 5O_2(g) \rightarrow 4CO_2(g) + 2H_2O(g) \)
4V \( CO_2 \) is collected with 2V \( C_2H_2 \)
So, 200cm³ \( CO_2 \) will be collected with = 100cm³ \( C_2H_2 \)
Similarly, 4V of \( CO_2 \) is produced by 5V of \( O_2 \)
So, 200cm³ \( CO_2 \) will be produced by = 250 ml of \( O_2 \)
In simple words: By knowing how much carbon dioxide is collected, we can work backwards to find how much acetylene and oxygen were used in the reaction.

📝 Teacher's Note: This is reverse stoichiometry - working from products back to reactants. Useful for experimental analysis problems.

🎯 Exam Tip: Use the mole ratios from the balanced equation to work backwards from CO₂ produced to find reactants consumed.

Exercise 5(B)

Solution 19.
Answer: This experiment supports Gay lussac's law of combining volumes. Since the unchanged or remaining O₂ is 58 cc so, used oxygen 106 – 58 = 48cc. According to Gay lussac's law, the volumes of gases reacting should be in a simple ratio.
\( CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \)
1 V 2 V
24 cc 48 cc
i.e. methane and oxygen react in a 1:2 ratio.
In simple words: When methane burns with oxygen, they combine in exact volume ratios - 1 part methane needs exactly 2 parts oxygen, proving Gay-Lussac's law about gas reactions.

📝 Teacher's Note: Use actual gas syringes or measuring cylinders to demonstrate this ratio practically. Students remember better when they see the 1:2 volume ratio physically.

🎯 Exam Tip: Always calculate used gas by subtracting remaining from initial amount. State the law name clearly and show the balanced equation with volume ratios.

Solution 19.
Answer: According to Avogadro's law, equal volumes of gases contain equal no. of molecules under similar conditions of temperature and pressure. This means more volume will contain more molecules and least volume will contain least molecules.
So,
(a) 5 litres of hydrogen has greatest no. of molecules with the maximum volume.
(b) 1 litre of SO₂ contains the least number of molecules since it has the smallest volume.
In simple words: Avogadro's law is like saying bigger containers hold more marbles - the gas with the largest volume has the most molecules.

📝 Teacher's Note: Compare this to balloons of different sizes - the bigger balloon has more air molecules inside. This visual analogy helps students understand volume-molecule relationship.

🎯 Exam Tip: State Avogadro's law first, then directly relate volume to number of molecules. Biggest volume = most molecules, smallest volume = least molecules.

Solution 20.
Answer:

In simple words: Using Avogadro's law, we can compare molecules by comparing volumes - nitrogen and ammonia have the same volume so same molecules, while chlorine and sulphur dioxide have smaller volumes so fewer molecules.

📝 Teacher's Note: Set nitrogen as the reference (x molecules) and calculate others as fractions or multiples of x. This builds proportional thinking skills.

🎯 Exam Tip: Choose one gas as reference and express all others in terms of that reference. Tables should be neat with clear headers and aligned data.

Solution 21.
Answer: 100 cm³ of oxygen contains = Y molecules
Applying Avogadro's law,
50 cm³ of nitrogen contains = \( \frac{50 Y}{100} = \frac{Y}{2} \)
In simple words: Since nitrogen has half the volume of oxygen, it contains half the number of molecules according to Avogadro's law.

📝 Teacher's Note: Emphasize that Avogadro's law creates direct proportionality - half volume means half molecules. Use simple fraction examples from daily life.

🎯 Exam Tip: Set up a simple proportion: Volume₁/Volume₂ = Molecules₁/Molecules₂. Cross multiply to solve for unknown molecules.

Solution 1.
Answer:
a) This statement means one atom of chlorine is 35.5 times heavier than 1/12 time of the mass of an atom C-12.
b) The value of avogadro's number is \( 6.023 \times 10^{23} \)
c) The molar volume of a gas at STP is 22.4 dm³ at STP
In simple words: These are three fundamental constants in chemistry - atomic mass comparison, number of particles in a mole, and volume occupied by gases at standard conditions.

📝 Teacher's Note: Connect these three concepts - atomic mass helps find molar mass, Avogadro's number counts particles, and molar volume measures gas space. They work together in calculations.

🎯 Exam Tip: Memorize these exact values: Avogadro's number = \( 6.023 \times 10^{23} \), molar volume = 22.4 dm³ at STP. State definitions precisely as given.

Solution 2.
Answer:
(a) The vapour density is the ratio between the masses of equal volumes of gas and hydrogen under the conditions of standard temperature and pressure.
(b) Molar volume is the volume occupied by one mole of the gas at STP. It is equal to 22.4 dm³.
(c) The relative atomic mass of an element is the number of times one atom of the element is heavier than 1/12 times of the mass of an atom of carbon-12.
(d) The relative molecular mass of an compound is the number that represents how many times one molecule of the substance is heavier than 1/12 of the mass of an atom of carbon-12.
(e) The number of atoms present in 12g (gram atomic mass) of C-12 isotope, i.e. \( 6.023 \times 10^{23} \) atoms.
(f) The quantity of the element which weighs equal to its gram atomic mass is called one gram atom of that element.
(g) Mole is the amount of a substance containing elementary particles like atoms, molecules or ions in 12 g of carbon-12.
In simple words: These definitions explain how we compare masses of atoms and molecules, count particles, and measure amounts of substances in chemistry.

📝 Teacher's Note: Create a concept map showing relationships between these terms. Start with carbon-12 as the reference point and build outward to other concepts.

🎯 Exam Tip: Learn the exact wording for definitions. Use "1/12 times the mass of carbon-12" phrase consistently. Always mention "equal volumes" for vapour density.

Solution 3.
Answer:
(a) Applications of Avogadro's Law :
1. It explains Gay-Lussac's law.
2. It determines atomicity of the gases.
3. It determines the molecular formula of a gas.
4. It determines the relation between molecular mass and vapour density.
5. It gives the relationship between gram molecular mass and gram molecular volume.
(b) According to Avogadro's law under the same conditions of temperature and pressure, equal volumes of different gases have the same number of molecules. Since substances react in simple ratio by number of molecules, volumes of the gaseous reactants and products will also bear a simple ratio to one another. This what Gay Lussac's Law says.
\( H_2 + Cl_2 \rightarrow 2HCl \)
1V 1V 2V (By Gay-Lussacs law)
n molecules n molecules 2n molecules (By Avogadros law)
In simple words: Avogadro's law helps us understand many gas behaviors and connects the number of molecules to volume, which explains why gases react in simple volume ratios.

📝 Teacher's Note: Show the connection between Avogadro's and Gay-Lussac's laws step by step. Use the same chemical equation to demonstrate both molecule counting and volume ratios.

🎯 Exam Tip: List applications in numbered points. For part (b), state both laws clearly and show how equal volumes containing equal molecules leads to simple volume ratios.

Solution 4.
Answer:
(a) (2N)28 + (8H)8 + (Pt)195 + (6Cl)35.5 × 6 = 444
(b) KClO₃ = (K)39 + (Cl)35.5 + (3O)48 = 122.5
(c) (Cu)63.5 + (S)32 + (4O)64 + (5H₂O)5 × 18 = 249.5
(d) (2N)28 + (8H)8 + (S)32 + (4O)64 = 132
(e) (C)12 + (3H)3 + (C)12 + (2O)32 + (Na)23 = 82
(f) (C)12 + (H)1+ (3Cl)3 × 35.5 = 119.5
(g) (2N)28 + (8H)8 + (2Cr)2 × 51.9+ (7O)7 × 16 = 252
In simple words: To find molecular mass, multiply the number of each type of atom by its atomic mass, then add all values together.

📝 Teacher's Note: Teach systematic approach: write formula, identify each element with subscript, multiply by atomic mass, add totals. Check calculations twice for accuracy.

🎯 Exam Tip: Show working clearly with element symbols, subscripts, and atomic masses. Double-check arithmetic - molecular mass questions are easy marks if calculated correctly.

Solution 5.
Answer:
(a) No. of molecules in 73 g HCl = \( 6.023 \times 10^{23} \times \frac{73}{36.5} \) (mol. mass of HCl) = \( 12.04 \times 10^{23} \)
(b) Weight of 0.5 mole of O₂ is = 32(mol. Mass of O₂) × 0.5=16 g
(c) No. of molecules in 1.8 g H₂O = \( 6.023 \times 10^{23} \times \frac{1.8}{18} = 6.023 \times 10^{22} \)
In simple words: Use the formula: molecules = Avogadro's number × (given mass ÷ molecular mass). For mass calculations, multiply moles by molecular mass.

📝 Teacher's Note: Practice the mole formula triangle: Mass at top, Moles and Molecular mass at bottom. Cover what you want to find and the formula appears.

🎯 Exam Tip: Always write the formula first: n = m/M or molecules = n × Nₐ. Substitute values carefully and show units in final answers.

Solution 6.
Answer: Molecular mass of H₂O is 18, CO₂ is 44, NH₃ is 17 and CO is 28. So, the weight of 1 mole of CO₂ is more than the other three.
In simple words: Among equal numbers of molecules (1 mole each), CO₂ molecules are heaviest, so 1 mole of CO₂ weighs the most.

📝 Teacher's Note: Clarify that 1 mole means same number of molecules, but different total masses due to different molecular weights. Use the analogy of counting different objects.

🎯 Exam Tip: Calculate or recall molecular masses for comparison. 1 mole mass = molecular mass in grams. Choose the highest molecular mass.

Solution 7.
Answer: 4g of NH₃ having minimum molecular mass contain maximum molecules.
In simple words: When you have the same mass of different substances, the one with smaller molecular mass will have more molecules - like having more small coins versus fewer large coins for the same total weight.

📝 Teacher's Note: Use the coin analogy - 100g of 1-rupee coins vs 100g of 10-rupee coins. Same weight, but more 1-rupee coins because each coin is lighter.

🎯 Exam Tip: For fixed mass, smaller molecular mass means more molecules. Use formula: molecules ∝ 1/molecular mass when mass is constant.

Solution 8.
Answer:
a) No. of particles in 1 mole = \( 6.023 \times 10^{23} \)
So, particles in 0.1 mole = \( 6.023 \times 10^{23} \times 0.1 = 6.023 \times 10^{22} \)
b) 1 mole of H₂SO₄ contains =2 × \( 6.023 \times 10^{23} \)
So, 0.1 mole of H₂SO₄ contains =2 × \( 6.023 \times 10^{23} \times 0.1 = 1.2 \times 10^{23} \) atoms of hydrogen
c) 111g CaCl₂ contains = \( 6.023 \times 10^{23} \) molecules
So, 1000 g contains = \( 5.42 \times 10^{24} \) molecules
In simple words: Multiply Avogadro's number by the number of moles to get total particles. For atoms within molecules, multiply by the number of that type of atom per molecule.

📝 Teacher's Note: For part (b), emphasize that H₂SO₄ has 2 hydrogen atoms per molecule, so multiply by 2. Draw the molecular structure to show atom count clearly.

🎯 Exam Tip: Read questions carefully - distinguish between molecules and specific atoms. For hydrogen atoms in H₂SO₄, multiply by 2 since there are 2 H atoms per molecule.

Solution 9.
Answer:
(a) 1 mole of aluminium has mass = 27 g
So, 0.2 mole of aluminium has mass = 0.2 × 27 = 5.4 g
(b) 0.1 mole of HCl has mass = 0.1 × 36.5 (mass of 1 mole) = 3.65 g
(c) 0.2 mole of H₂O has mass = 0.2 × 18 = 3.6 g
(d) 0.1 mole of CO₂ has mass = 0.1 × 44 = 4.4 g
In simple words: To find mass from moles, multiply the number of moles by the molar mass (atomic or molecular mass in grams).

📝 Teacher's Note: Use the mole triangle consistently. Practice with different substances so students become fluent with the mass = moles × molar mass relationship.

🎯 Exam Tip: Formula is Mass = moles × molar mass. Always include units (grams) in your final answer and show the multiplication clearly.

Solution 10.
Answer:
(a) 5.6 litres of gas at STP has mass = 12 g
So, 22.4 litre (molar volume) has mass =12 × 22.4/5.6 = 48g(molar mass)
(b) 1 mole of SO₂ has volume = 22.4 litres
So, 2 moles will have = 22.4 × 2 = 44.8 litre
In simple words: Use proportions to find molar mass from given volume and mass. For volume calculations, multiply moles by 22.4 L at STP.

📝 Teacher's Note: Emphasize that molar volume (22.4 L) applies only at STP conditions. Use cross-multiplication method for proportion problems.

🎯 Exam Tip: Set up proportions: if 5.6 L has 12g mass, then 22.4 L has x grams. Cross multiply to solve for x (molar mass).

Solution 11.
Answer:
(a) 1 mole of CO₂ contains O₂ = 32g
So, CO₂ having 8 gm of O₂ has no. of moles = 8/32 = 0.25 moles
(b) 16 g of methane has no. of moles = 1
So, 0.80 g of methane has no. of moles = 0.8/16 = 0.05 moles
In simple words: Find what fraction of the molar amount you have by dividing given amount by the molar amount.

📝 Teacher's Note: For part (a), clarify that CO₂ contains oxygen atoms, and we're finding moles of CO₂ based on oxygen content. Draw the molecule structure.

🎯 Exam Tip: Use moles = given mass ÷ molar mass. For CO₂, if asked about oxygen content, remember CO₂ has 32g of oxygen per mole.

Solution 12.
Answer:
(a) \( 6.023 \times 10^{23} \) atoms of oxygen has mass = 16 g
So, 1 atom has mass = \( \frac{16}{6.023 \times 10^{23}} = 2.656 \times 10^{-23} \) g
(b) 1 atom of Hydrogen has mass = \( \frac{1}{6.023 \times 10^{23}} = 1.666 \times 10^{-24} \)
(c) 1 molecule of NH₃ has mass = \( \frac{17}{6.023 \times 10^{23}} = 2.82 \times 10^{-23} \) g
(d) 1 atom of silver has mass = \( \frac{108}{6.023 \times 10^{23}} = 1.701 \times 10^{-22} \)
(e) 1 molecule of O₂ has mass = \( \frac{32}{6.023 \times 10^{23}} = 5.314 \times 10^{-23} \) g
(f) 0.25 gram atom of calcium has mass = 0.25 × 40 = 10g
In simple words: To find mass of single atoms or molecules, divide the molar mass by Avogadro's number. The numbers become very small since atoms are extremely tiny.

📝 Teacher's Note: Emphasize how incredibly small individual atoms are. Use scientific notation properly and explain why masses are so tiny with negative exponents.

🎯 Exam Tip: Formula is: mass of 1 particle = molar mass ÷ Avogadro's number. Express answers in scientific notation with proper significant figures.

Solution 13.
Answer:
(a) 0.1 mole of CaCO₃ has mass =100(molar mass) × 0.1=10 g
(b) 0.1 mole of Na₂SO₄.10H₂O has mass = 322 × 0.1 = 32.2 g
(c) 0.1 mole of CaCl₂ has mass = 111 × 0.1 = 11.1g
(d) 0.1 mole of Mg has mass = 24 × 0.1 = 2.4 g
In simple words: Multiply the number of moles by the molar mass to get the total mass. For hydrated salts, include the water molecules in molar mass calculation.

📝 Teacher's Note: For hydrated salts like Na₂SO₄.10H₂O, teach students to add the mass of 10 water molecules (180g) to the anhydrous salt mass (142g).

🎯 Exam Tip: Calculate molar mass correctly first, especially for hydrated compounds. Then multiply by moles. Show the molar mass calculation clearly.

Solution 14.
Answer: 1molecule of Na₂CO₃.10H₂O contains oxygen atoms = 13
So, \( 6.023 \times 10^{23} \) molecules (1mole) has atoms=13 × \( 6.023 \times 10^{23} \)
So, 0.1 mole will have atoms = 0.1 × 13 × \( 6.023 \times 10^{23} = 7.8 \times 10^{23} \)
In simple words: Count oxygen atoms in the formula (3 in carbonate + 10 in water = 13), then multiply by number of molecules present.

📝 Teacher's Note: Break down the formula: Na₂CO₃ has 3 oxygen atoms, 10H₂O has 10 oxygen atoms, total = 13. Draw the structure to show atom counting.

🎯 Exam Tip: First count atoms per molecule carefully from the chemical formula. Then multiply by total number of molecules (moles × Avogadro's number).

Solution 15.
Answer: 3.2 g of S has number of atoms = \( 6.023 \times 10^{23} \times \frac{3.2}{32} = 0.6023 \times 10^{23} \)
In simple words: Find how many moles you have (3.2g ÷ 32g/mol = 0.1 mol), then multiply by Avogadro's number to get total atoms.

📝 Teacher's Note: Show the two-step process: first find moles, then find atoms. This reinforces the relationship between mass, moles, and particle count.

🎯 Exam Tip: Use the formula: atoms = (given mass ÷ atomic mass) × Avogadro's number. Express final answer in proper scientific notation.

Answer: So, \( 0.6023 \times 10^{23} \) atoms of Ca has mass = \( 40 \times 0.6023 \times 10^{23}/6.023 \times 10^{23} = 4g \)

📝 Teacher's Note: Emphasize that Avogadro's number is the key conversion factor between atoms and grams. Show students how the powers of 10 cancel out neatly in calculations.

🎯 Exam Tip: Always write Avogadro's number as \( 6.023 \times 10^{23} \) and show the cancellation of units clearly to avoid calculation errors.

Solution 16.
Answer:
(a) No. of atoms = \( 52 \times 6.023 \times 10^{23} = 3.131 \times 10^{25} \)
(b) 4 amu = 1 atom of He
so, 52 amu = 13 atoms of He
(c) 4 g of He has atoms = \( 6.023 \times 10^{23} \)
So, 52 g will have = \( 6.023 \times 10^{23} \times 52/4 = 7.828 \times 10^{24} \) atoms

📝 Teacher's Note: Help students distinguish between atomic mass units (amu) and grams. Use simple ratios to show how mass relates to number of atoms in different units.

🎯 Exam Tip: Remember that 1 mole of any element has a mass in grams equal to its atomic mass in amu, and contains Avogadro's number of atoms.

Solution 17.
Answer:
Molecular mass of \( Na_2CO_3 = 106 g \)
106 g has \( 2 \times 6.023 \times 10^{23} \) atoms of Na
So, 5.3g will have = \( 2 \times 6.023 \times 10^{23} \times 5.3/106 = 6.022 \times 10^{22} \) atoms
Number of atoms of C = \( 6.023 \times 10^{23} \times 5.3/106 = 3.01 \times 10^{22} \) atoms
And atoms of O = \( 3 \times 6.023 \times 10^{23} \times 5.3/106 = 9.03 \times 10^{22} \) atoms

📝 Teacher's Note: Point out that in \( Na_2CO_3 \), there are 2 Na atoms, 1 C atom, and 3 O atoms per molecule. Students must multiply by these subscripts when calculating individual element atoms.

🎯 Exam Tip: Always identify the subscripts in the molecular formula first, then use them as multipliers when finding atoms of individual elements.

Solution 18.
Answer:
(a) 60 g urea has mass of nitrogen(\( N_2 \)) = 28 g
So, 5000 g urea will have mass = \( 28 \times 5000/60 = 2.33 kg \)
(b) 64 g has volume = 22.4 litre
So, 320 g will have volume = \( 22.4 \times 320/64 = 112 \) litres

📝 Teacher's Note: Explain that urea contains nitrogen and we're finding what mass of nitrogen is present in a given mass of urea using proportion method.

🎯 Exam Tip: Set up proportions carefully: if smaller mass gives smaller result, then larger mass gives proportionally larger result.

Solution 19.
Answer:
(a) Vapour density of carbon dioxide is 22, it means that 1 molecule of carbon dioxide is 22 times heavier than 1 molecule of hydrogen.
(b) Vapour density of Chlorine atom is 35.5.

📝 Teacher's Note: Emphasize that vapor density compares the mass of a gas molecule to hydrogen molecule (not hydrogen atom). This is a fundamental concept in gas calculations.

🎯 Exam Tip: Vapor density = Molecular mass ÷ 2, because hydrogen gas (\( H_2 \)) has molecular mass 2.

Solution 20.
Answer:
22400 \( cm^3 \) of CO has mass = 28 g
So, 56 \( cm^3 \) will have mass = \( 56 \times 28/22400 = 0.07 g \)

📝 Teacher's Note: Remind students that 22400 \( cm^3 \) = 22.4 L, which is the molar volume of any gas at STP. This is a key reference point for gas calculations.

🎯 Exam Tip: Remember that 1 mole of any gas occupies 22.4 L or 22400 \( cm^3 \) at STP.

Solution 21.
Answer:
18 g of water has number of molecules = \( 6.023 \times 10^{23} \)
So, 0.09 g of water will have no. of molecules = \( 6.023 \times 10^{23} \times 0.09/18 = 3.01 \times 10^{21} \) molecules

📝 Teacher's Note: Use water as a familiar example to show how we use the molar mass (18 g/mol for \( H_2O \)) to convert between grams and number of molecules.

🎯 Exam Tip: Always start with the molar mass of the compound to set up your proportion correctly.

Solution 22.
Answer:
(a) No. of moles in 256 g \( S_8 = 1 \) mole
So, no. of moles in 5.12 g = \( 5.12/256 = 0.02 \) moles
(b) No. of molecules = \( 0.02 \times 6.023 \times 10^{23} = 1.2 \times 10^{22} \) molecules
No. of atoms in 1 molecule of S = 8
So, no. of atoms in \( 1.2 \times 10^{22} \) molecules = \( 1.2 \times 10^{22} \times 8 = 9.635 \times 10^{22} \) molecules

📝 Teacher's Note: Point out that sulfur exists as \( S_8 \) molecules, so each molecule contains 8 sulfur atoms. This is different from elements that exist as single atoms.

🎯 Exam Tip: When finding total atoms, multiply the number of molecules by the number of atoms per molecule (the subscript in the formula).

Solution 23.
Answer:
Atomic mass of phosphorus P = 30.97 g
Hence, molar mass of \( P_4 = 123.88 g \)
If phosphorus is considered as \( P_4 \) molecules,
then 1 mole \( P_4 \equiv 123.88 g \)
Therefore, 100 g of \( P_4 = 0.807 \) moles

📝 Teacher's Note: Explain that some elements like phosphorus exist as molecules (\( P_4 \)) rather than individual atoms, so we must use the molecular mass for calculations.

🎯 Exam Tip: Check whether the element exists as single atoms or molecules - this affects which molar mass to use in calculations.

Solution 24.
Answer:
(a) 308 \( cm^3 \) of chlorine weighs = 0.979 g
So, 22400 \( cm^3 \) will weigh = gram molecular mass = \( 0.979 \times 22400/308 = 71.2 g \)
(b) 2 g(molar mass) \( H_2 \) at 1 atm has volume = 22.4 litres
So, 4 g \( H_2 \) at 1 atm will have volume = 44.8 litres
Now, at 1 atm(\( P_1 \)) 4 g \( H_2 \) has volume (\( V_1 \)) = 44.8 litres
So, at 4 atm(\( P_2 \)) the volume(\( V_2 \)) will be = \( \frac{P_1V_1}{P_2} = \frac{1 \times 44.8}{4} = 11.2 \) litres
(c) Mass of oxygen in 22.4 litres = 32 g(molar mass)
So, mass of oxygen in 2.2 litres = \( 2.2 \times 32/22.4 = 3.14 g \)

📝 Teacher's Note: This problem combines gas laws with molar calculations. Show students how to use Boyle's law (\( P_1V_1 = P_2V_2 \)) alongside molar volume concepts.

🎯 Exam Tip: Break complex problems into steps: first find molar mass or volume at STP, then apply gas laws if conditions change.

Solution 25.
Answer:
No. of atoms in 12 g C = \( 6.023 \times 10^{23} \)
So, no. of carbon atoms in \( 10^{-12} g = 10^{-12} \times 6.023 \times 10^{23}/12 = 5.019 \times 10^{10} \) atoms

📝 Teacher's Note: This involves very small masses and large numbers. Help students handle scientific notation carefully, especially when dividing powers of 10.

🎯 Exam Tip: When dealing with scientific notation, add or subtract exponents carefully. \( 10^{-12} \times 10^{23} = 10^{11} \).

Solution 26.
Answer:
Given:
P = 1140 mm Hg
Density = D = 2.4 g / L
T = 273°C = 273+273 = 546 K
M = ?

We know that, at STP, the volume of one mole of any gas is 22.4 L
Hence we have to find out the volume of the unknown gas at STP.

First apply Charles's law.
We have to find out the volume of one liter of unknown gas at standard temperature 273 K.
V₁ = 1 L, T₁ = 546 K
V₂ = ?, T₂ = 273 K
V₁/T₁ = V₂/T₂
V₂ = (V₁ × T₂)/T₁ = (1 L × 273 K)/546 K = 0.5 L

We have found out the volume at standard temperature. Now we have to find out the volume at standard pressure.
Apply Boyle's law.
P₁ = 1140 mm Hg, V₁ = 0.5 L
P₂ = 760 mm Hg, V₂ = ?
P₁ × V₁ = P₂ × V₂
V₂ = (P₁ × V₁)/P₂ = (1140 mm Hg × 0.5 L)/760 mm Hg = 0.75 L

Now, 22.4 L is the volume of 1 mole of any gas at STP, then 0.75 L is the volume of X moles at STP
X moles = 0.75 L / 22.4 L = 0.0335 moles
The original mass is 2.4 g
n = m / M
0.0335 moles = 2.4 g / M
M = 2.4 g / 0.0335 moles = 71.6 g / mole
Hence, the gram molecular mass of the unknown gas is 71.6 g

📝 Teacher's Note: This is a comprehensive gas law problem. Teach students to work systematically: first convert to STP conditions using gas laws, then use molar volume to find molecular mass.

🎯 Exam Tip: Always convert to STP first (273 K, 760 mm Hg) before using the 22.4 L molar volume. Apply Charles's law for temperature, then Boyle's law for pressure.

Solution 27.
Answer:
1000 g of sugar costs = Rs. 40
So, 342g(molar mass) of sugar will cost = \( 342 \times 40/1000 = Rs. 13.68 \)

📝 Teacher's Note: This connects chemistry with everyday life. The molar mass of sugar (sucrose, \( C_{12}H_{22}O_{11} \)) is 342 g, which represents one mole of sugar molecules.

🎯 Exam Tip: Use simple proportion: if 1000 g costs Rs. 40, then 342 g costs proportionally less.

Solution 28.
Answer:
(a) Weight of 1 g atom N = 14 g
So, weight of 2 g atom of N = 28 g
(b) \( 6.023 \times 10^{23} \) atoms of C weigh = 12 g
So, \( 3 \times 10^{25} \) atoms will weigh = \( \frac{12 \times 3 \times 10^{25}}{6.023 \times 10^{23}} = 597.7 g \)
(c) 1 mole of sulphur weighs = 32 g
(d) 7 g of silver
So, 7 grams of silver weighs least.

📝 Teacher's Note: Help students understand that "g atom" means "gram atomic mass" or one mole of atoms. Compare different quantities to find which has the smallest mass.

🎯 Exam Tip: Convert all quantities to grams, then compare. Remember that 1 g atom = 1 mole = atomic mass in grams.

Solution 29.
Answer:
40 g of NaOH contains \( 6.023 \times 10^{23} \) molecules
So, 4 g of NaOH contains = \( 6.02 \times 10^{23} \times 4/40 = 6.02 \times 10^{22} \) molecules

📝 Teacher's Note: The molar mass of NaOH is 40 g, so this is a direct proportion problem. One-tenth the mass means one-tenth the molecules.

🎯 Exam Tip: Use the molar mass as your reference point: 40 g NaOH = 1 mole = Avogadro's number of molecules.

Solution 30.
Answer:
The number of molecules in 18 g of ammonia = \( 6.02 \times 10^{23} \)
So, no. of molecules in 4.25 g of ammonia = \( 6.02 \times 10^{23} \times 4.25/18 = 1.5 \times 10^{23} \)

📝 Teacher's Note: The molar mass of ammonia (\( NH_3 \)) is approximately 17 g, but the problem uses 18 g for calculation. Students should use the given values in problems.

🎯 Exam Tip: Always use the values provided in the problem, even if they differ slightly from calculated molar masses.

Solution 31.
Answer:
(a) One mole of chlorine contains \( 6.023 \times 10^{23} \) atoms of chlorine.
(b) Under similar conditions of temperature and pressure, two volumes of hydrogen combined with one volume of oxygen will give two volumes of water vapour.
(c) Relative atomic mass of an element is the number of times one atom of an element is heavier than 1/12 the mass of an atom of carbon-12.
(d) Under similar conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules.

📝 Teacher's Note: These are fundamental definitions in chemistry. Statement (b) is Gay-Lussac's law, (c) defines relative atomic mass, and (d) is Avogadro's law.

🎯 Exam Tip: Learn these definitions by heart - they form the foundation for solving numerical problems in chemistry.

Exercise 5(C)

Solution 1.
Answer:
Information conveyed by \( H_2O \)
1. That \( H_2O \) contains 2 volumes of hydrogen and 1 volume of oxygen.
2. That ratio by weight of hydrogen and oxygen is 1:8.
3. That molecular weight of \( H_2O \) is 18g.

📝 Teacher's Note: A molecular formula gives multiple types of information: volume ratios (for gases), mass ratios, and absolute molecular mass. Use water as a familiar example.

🎯 Exam Tip: Remember that molecular formulas tell us both qualitative (which elements) and quantitative (how many atoms, mass ratios) information.

Solution 2.
Answer:
The empirical formula is the simplest formula, which gives the simplest ratio in whole numbers of atoms of different elements present in one molecule of the compound.
The molecular formula of a compound denotes the actual number of atoms of different elements present in one molecule of a compound.

📝 Teacher's Note: Use examples like glucose (\( C_6H_{12}O_6 \) molecular, \( CH_2O \) empirical) to show the difference. The empirical formula is like a simplified recipe ratio.

🎯 Exam Tip: Empirical formula = simplest whole number ratio; Molecular formula = actual number of atoms in one molecule.

Solution 3.
Answer:
(a) CH (b) \( CH_2O \) (c) CH (d) \( CH_2O \)

📝 Teacher's Note: Students need to practice finding the simplest whole number ratios from given molecular formulas by dividing by the greatest common factor.

🎯 Exam Tip: To find empirical formula from molecular formula, divide all subscripts by their greatest common divisor.

Solution 4.
Answer:
Relative mol. mass of \( CuSO_4.5H_2O = 63.5+32+(16 \times 4)+5(1 \times 2+16) = 249.5 g \)
249.5 g of \( CuSO_4.5H_2O \) contains water of crystallization = 90 g
So, 100 g will contain = \( \frac{90 \times 100}{249.5} = 36.07 g \)
So, % of \( H_2O = 36.07 \times 100 = 36.07% \)

📝 Teacher's Note: This is a hydrated salt calculation. Show students how to calculate the mass of water molecules separately (5 × 18 = 90 g) then find its percentage of the total mass.

🎯 Exam Tip: For hydrated salts, calculate the mass contribution of water molecules separately, then find what percentage they represent of the total molar mass.

Solution 5.
Answer:
(a) Molecular mass of \( Ca(H_2PO_4)_2 = 234 \)
So, % of P = \( 2 \times 31 \times 100/234 = 26.5% \)
(b) Molecular mass of \( Ca_3(PO_4)_2 = 310 \)
% of P = \( 2 \times 31 \times 100/310 = 20% \)

📝 Teacher's Note: Point out that both compounds contain phosphorus but in different amounts. The subscripts and parentheses must be handled carefully in molecular mass calculations.

🎯 Exam Tip: Count atoms carefully when parentheses are present: \( Ca(H_2PO_4)_2 \) has 2 P atoms, \( Ca_3(PO_4)_2 \) also has 2 P atoms.

Solution 6.
Answer:
Molecular mass of \( KClO_3 = 122.5 g \)
% of K = \( 39/122.5 = 31.8% \)
% of Cl = \( 35.5/122.5 = 28.98% \)
% of O = \( 3 \times 16/122.5 = 39.18% \)

📝 Teacher's Note: This is a straightforward percentage composition calculation. Remind students that the sum of all percentages should equal 100% (allowing for rounding errors).

🎯 Exam Tip: Always check that your percentages add up to approximately 100% - this catches calculation errors quickly.

Solution 7.
Answer: Element % At. mass Atomic ratio Simple ratio
Pb 62.5 207 \( \frac{62.5}{207} = 0.3019^1 \)
N 8.5 14 \( \frac{8.5}{14} = 0.6071^2 \)
O 29.0 16 \( \frac{29.0}{16} = 1.81^6 \)
So, Pb(NO₃)₂ is the empirical formula.

📝 Teacher's Note: Always divide the percentage by atomic mass first, then find the simplest ratio by dividing all values by the smallest one. Students often forget to round properly to get whole numbers.

🎯 Exam Tip: Show all calculation steps clearly and write the final empirical formula in proper chemical notation with subscripts.

Solution 8.
Answer: In Fe₂O₃, Fe = 56 and O = 16
Molecular mass of Fe₂O₃ = 2 × 56 + 3 × 16 = 160 g
Iron present in 80% of Fe₂O₃ = \( \frac{112}{160} \) × 80 = 56 g
So, mass of iron in 100 g of ore = 56 g
∴ mass of Fe in 10000 g of ore = 56 × 10000/100 = 5.6 kg

📝 Teacher's Note: Emphasize that students must first find the mass of iron in the pure compound, then apply the percentage purity of the ore.

🎯 Exam Tip: Always convert to the required units at the end - here grams to kilograms. Double-check unit conversions.

Solution 9.
Answer: For acetylene, molecular mass = 2 × V.D = 2 × 13 = 26 g
The empirical mass = 12(C) + 1(H) = 13 g
n = \( \frac{\text{Molecular formula mass}}{\text{Empirical formula weight}} = \frac{26}{13} = 2 \)
Molecular formula of acetylene = 2 × Empirical formula = C₂H₂
Similarly, for benzene molecular mass = 2 × V.D = 2 × 39 = 78
n = 78/13 = 6
So, the molecular formula = C₆H₆

📝 Teacher's Note: Remind students that vapour density is always half the molecular mass, so molecular mass = 2 × V.D. This is a key relationship.

🎯 Exam Tip: Always calculate the value of 'n' first, then multiply the empirical formula by n to get the molecular formula.

Solution 10.
Answer: Element % At. mass Atomic ratio Simple ratio
H 17.7 1 \( \frac{17.7}{1} = 17.7 \) \( \frac{17.7}{5.87} = 3 \)
N 82.3 14 \( \frac{82.3}{14} = 5.87 \) \( \frac{5.87}{5.87} = 1 \)
So, the empirical formula = NH₃

📝 Teacher's Note: Students should always divide by the smallest atomic ratio to get the simplest whole number ratio. Show them how 17.7/5.87 ≈ 3.

🎯 Exam Tip: When ratios aren't perfect whole numbers, round to the nearest whole number if the decimal is close to 0.5 or above.

Solution 11.
Answer: Element % at. mass atomic ratio simple ratio
C 54.54 12 \( \frac{54.54}{12} = 4.55^2 \)
H 9.09 1 \( \frac{9.09}{1} = 9.09^4 \)
O 36.36 16 \( \frac{36.36}{16} = 2.27^1 \)
(a) So, its empirical formula = C₂H₄O
(b) empirical formula mass = 44
Since, vapour density = 44
So, molecular mass = 2 × V.D = 88
Or n = 2
so, molecular formula = (C₂H₄O)₂ = C₄H₈O₂

📝 Teacher's Note: Show students how to use vapour density to find molecular mass, then calculate n = molecular mass/empirical formula mass.

🎯 Exam Tip: Remember that molecular mass = 2 × vapour density. This relationship is frequently tested.

Solution 12.
Answer: Element % at. mass atomic ratio simple ratio
C 26.59 12 \( \frac{26.59}{12} = 2.21^1 \)
H 2.22 1 \( \frac{2.22}{1} = 2.22^1 \)
O 71.19 16 \( \frac{71.19}{16} = 4.44^2 \)
(a) its empirical formula = CHO₂
(b) empirical formula mass = 45
Vapour density = 45
So, molecular mass = V.D × 2 = 90
so, molecular formula = C₂H₂O₄

📝 Teacher's Note: Point out that when all atomic ratios are approximately the same, the empirical formula has a 1:1:2 ratio after simplification.

🎯 Exam Tip: Check your empirical formula mass calculation - it should match the given vapour density when n=1, or be a simple multiple of it.

Solution 13.
Answer: Element % at. mass atomic ratio simple ratio
Cl 71.65 35.5 \( \frac{71.65}{35.5} = 2.01^1 \)
H 4.07 1 \( \frac{4.07}{1} = 4.07^2 \)
C 24.28 12 \( \frac{24.28}{12} = 2.02^1 \)
(a) its empirical formula = CH₂Cl
(b) empirical formula mass = 49.5
Since, molecular mass = 98.96
so, molecular formula = (CH₂Cl)₂ = C₂H₄Cl₂

📝 Teacher's Note: When ratios are very close to whole numbers (like 2.01 ≈ 2), round them to the nearest integer. This is normal due to experimental error.

🎯 Exam Tip: Always write both empirical and molecular formulas when asked, and show the calculation of n clearly.

Solution 14.
Answer: (a) The g atom of carbon = 4.8/12 = 0.4 and g atom of hydrogen = 1/1=1
(b) Element Given mass At. mass Gram atom Ratio
C 4.8 12 0.4 1
H 1 1 1 2.5
So, the empirical formula = C₂H₅
(c) Empirical formula mass = 29
Molecular mass = V.D × 2 = 29 × 2 = 58
So, molecular formula = C₄H₁₀

📝 Teacher's Note: When you get a ratio like 1:2.5, multiply both by 2 to get 2:5, giving the empirical formula C₂H₅.

🎯 Exam Tip: If ratios contain decimals like 2.5, multiply all ratios by the appropriate number to get whole numbers.

Solution 15.
Answer: Since, g atom of Si = given mass/mol. Mass
so, given mass = 0.2 × 28 = 5.6 g
Element mass At. mass Gram atom Ratio
Si 5.6 28 0.2 1
Cl 21.3 35.5 \( \frac{21.3}{35.5} = 0.6^3 \)
Empirical formula = SiCl₃

📝 Teacher's Note: Students should understand that gram atoms = given mass ÷ atomic mass. This is the foundation for finding empirical formulas.

🎯 Exam Tip: When given gram atoms directly, use them to find the given mass first, then proceed with the ratio calculation.

Solution 16.
Answer: Element % at. mass atomic ratio simple ratio
C 92.3 12 \( \frac{92.3}{12} = 7.7^1 \)
H 7.7 1 \( \frac{7.7}{1} = 7.7^1 \)
So, empirical formula is CH
Empirical formula mass = 13
Since molecular mass = 78
So, n = 6
∴ molecular formula is C₆H₆

📝 Teacher's Note: This is benzene - a classic example where the empirical formula (CH) is much simpler than the molecular formula (C₆H₆).

🎯 Exam Tip: When the empirical formula seems too simple (like CH), always calculate n to find the actual molecular formula.

Solution 17.
Answer: (a) G atoms of magnesium = 18/24 = 0.75 or g-atom of Mg
(b) G atoms of nitrogen = 7/14 = 0.5 or 1/2 g-atoms of N
(c) Ratio of gram-atoms of N and Mg = 1:1.5 or 2:3
So, the formula is Mg₃N₂

📝 Teacher's Note: When ratios are not whole numbers (like 1:1.5), multiply by the appropriate factor to get the simplest whole number ratio (2:3).

🎯 Exam Tip: Always express the final formula with the metal first, followed by the non-metal, following standard chemical notation.

Solution 18.
Answer: Barium chloride = BaCl₂.x H₂O
Ba + 2Cl + x[H₂ + O]
= 137+ 2×35.5 + x [2+16]
= [208 + 18x] contains water = 14.8% water in BaCl₂.x H₂O
= [208 + 18 x] 14.8/100 = 18x
= [104 + 9x] 2148=18000x
= [104+9x] 37=250x
= 3848 + 333x =2250x
1917x =3848
x = 2molecules of water

📝 Teacher's Note: This problem involves setting up equations based on percentage composition. The key is that the water molecules contribute both to mass and to the percentage.

🎯 Exam Tip: When dealing with hydrated salts, set up the equation carefully: (mass of water/total mass) × 100 = given percentage.

Solution 19.
Answer: Molar mass of urea; CON₂H₄= 60 g
So, % of Nitrogen = 28 × 100/60 = 46.66%

📝 Teacher's Note: For percentage composition, the formula is: (mass of element × 100) ÷ molecular mass. In urea, there are 2 nitrogen atoms, so mass = 2 × 14 = 28.

🎯 Exam Tip: Count the number of atoms of each element carefully - urea has 2 nitrogen atoms, not 1.

Solution 20.
Answer: Element % At. mass Atomic ratio Simple ratio
C 42.1 12 3.5 1
H 6.48 1 6.48 2
O 51.42 16 3.2 1
The empirical formula is CH₂O
Since the compound has 12 atoms of carbon, so the formula is C₁₂H₂₄O₁₂.

📝 Teacher's Note: When told the compound has a specific number of atoms (like 12 carbons), multiply the empirical formula by the appropriate factor.

🎯 Exam Tip: Read the question carefully - if they specify the number of atoms of any element, use that to determine the molecular formula directly.

Solution 21.
Answer: (a) Now since the empirical formula is equal to vapour density and we know that vapour density is half of the molecular mass i.e. we have n=2 so, molecular formula is A₂B₄.
(b) Since molecular mass is 2 times the vapour density, so Mol. Mass = 2 V.D
Empirical formula weight = V.D/3
So, n = molecular mass/ Empirical formula weight = 6
Hence, the molecular formula is A₆B₆

📝 Teacher's Note: This is a conceptual problem testing understanding of the relationship between empirical formulas, molecular mass, and vapour density.

🎯 Exam Tip: Remember the key relationships: molecular mass = 2 × V.D, and n = molecular mass ÷ empirical formula mass.

Solution 22.
Answer: Atomic ratio of N = 87.5/14 = 6.25
Atomic ratio of H= 12.5/1 = 12.5
This gives us the simplest ratio as 1:2
So, the molecular formula is NH₂

📝 Teacher's Note: To get the simplest ratio, divide both atomic ratios by the smallest one: 6.25/6.25 = 1 and 12.5/6.25 = 2.

🎯 Exam Tip: Always simplify ratios to the smallest possible whole numbers - here 6.25:12.5 becomes 1:2.

Solution 23.
Answer: Element % at. mass atomic ratio simple ratio
Zn 22.65 65 0.348 1
H 4.88 1 4.88 14
S 11.15 32 0.348 1
O 61.32 16 3.83 11
Empirical formula of the given compound =ZnSH₁₄O₁₁
Empirical formula mass = 65.37+32+14×1+11×16=287.37
Molecular mass = 287
n = Molecular mass/Empirical formula mass = 287/287=1
Molecular formula = ZnSO₁₁H₁₄
= ZnSO₄.7H₂O

📝 Teacher's Note: This compound is zinc sulfate heptahydrate. The final step involves recognizing that ZnSO₄ is the main compound with 7 water molecules attached.

🎯 Exam Tip: For hydrated salts, group the atoms logically: identify the main salt first, then the water molecules (every 2H + 1O = 1 H₂O).

Exercise 5(D)

Solution 1.
Answer: (a) Moles:1 mole + 2 mole → 1 mole + 2 mole
(b) Grams: 42g + 36g → 74g + 4 g
(c) Molecules = 6.02 × 10²³ + 12.046 × 10²³ → 6.02 × 10²³+ 12.046 × 10²³

📝 Teacher's Note: This demonstrates the law of conservation of mass - mass and moles are conserved in chemical reactions, but volume may change.

🎯 Exam Tip: Always balance equations first, then apply stoichiometric relationships. Mass before = Mass after in any chemical reaction.

Solution 2.
Answer: (a) 100 g of CaCO₃ produces = 164 g of Ca(NO₃)₂
So, 15 g CaCO₃ will produce = 164 × 15/100 = 24.6 g Ca(NO₃)₂
(b) 1 V of CaCO₃ produces 1 V of CO₂
100 g of CaCO₃ has volume = 22.4 litres
So, 15 g will have volume = 22.4 × 15/100 = 3.36 litres CO₂

📝 Teacher's Note: Use molar ratios from balanced equations. 1 mole CaCO₃ produces 1 mole CO₂, so volumes are equal at STP.

🎯 Exam Tip: Remember that 1 mole of any gas occupies 22.4 L at STP. Use this to convert between mass and volume of gases.

Solution 3.
Answer: \( 2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4 \)
66 g
(a) \( 2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4 \)
34 g 98 g 132 g
For 132 g \( (NH_4)_2SO_4 \) = 34 g of \( NH_3 \) is required
So, for 66 g \( (NH_4)_2SO_4 \) = 66 × 32/132 = 17 g of \( NH_3 \) is required
(b) 17 g of \( NH_3 \) requires volume = 22.4 litres
(c) Mass of acid required, for producing 132g \( (NH_4)_2SO_4 \) = 98g
So, Mass of acid required, for 66g \( (NH_4)_2SO_4 \) = 66 × 98/132 = 49g
In simple words: We use stoichiometry to find how much ammonia, volume, and acid is needed to make a certain amount of ammonium sulfate. Just like following a recipe where you need exact amounts of each ingredient.

📝 Teacher's Note: Start with the balanced equation and teach students to set up proportions using molar masses. Emphasize that all calculations come from the balanced chemical equation.

🎯 Exam Tip: Always write the balanced equation first, then set up proportions clearly showing units. This prevents calculation errors and shows your method.

Solution 4.
Answer: (a) Molecular mass of \( Pb_3O_4 \) = 3 × 207.2 + 4 × 16 = 685 g
685 g of \( Pb_3O_4 \) gives = 834 g of \( PbCl_2 \)
Hence, 6.85 g of \( Pb_3O_4 \) will give = 6.85 × 834/685 = 8.34 g
(b) 685g of \( Pb_3O_4 \) gives = 71g of \( Cl_2 \)
Hence, 6.85 g of \( Pb_3O_4 \) will give = 6.85 × 71/685 = 0.71 g \( Cl_2 \)
(c) 1 V \( Pb_3O_4 \) produces 1 V \( Cl_2 \)
685g of \( Pb_3O_4 \) has volume = 22.4 litres = volume of \( Cl_2 \) produced
So, 6.85 \( Pb_3O_4 \) will produce = 6.85 × 22.4/685 = 0.224 litres of \( Cl_2 \)
In simple words: When lead oxide reacts, we can calculate how much lead chloride and chlorine gas will form using simple ratios based on the chemical equation.

📝 Teacher's Note: Show students how to identify what's being asked (mass vs volume) and use appropriate conversion factors. Practice with simpler numbers first.

🎯 Exam Tip: For gas volume calculations, remember that 1 mole of any gas at STP = 22.4 L. Always check if the question asks for mass or volume.

Solution 5.
Answer: Molecular mass of \( KNO_3 \) = 101 g
63 g of \( HNO_3 \) is formed by = 101 g of \( KNO_3 \)
So, 126000 g of \( HNO_3 \) is formed by = 126000 × 101/63 = 202 kg
Similarly, 126 g of \( HNO_3 \) is formed by 170 kg of \( NaNO_3 \)
So, smaller mass of \( NaNO_3 \) is required.
In simple words: We're comparing two different salts to see which one needs less mass to make the same amount of nitric acid. Sodium nitrate is more efficient.

📝 Teacher's Note: This is a great problem to show students how to compare different starting materials for the same product. Emphasize the importance of molecular masses in such comparisons.

🎯 Exam Tip: When comparing efficiency of reactants, calculate the mass needed from each and choose the smaller value. Always state your conclusion clearly.

Solution 6.
Answer: \( CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2 \)
100g 73g 22.4L
(a) \( V_1 \) = 2 litres \( V_2 \) = ?
\( T_1 \) = (273+27) = 300K \( T_2 \) = 273K
\( V_1/T_1 = V_2/T_2 \)
\( V_2 = V_1T_2/T_1 = \frac{2 \times 273}{300} \)L
Now at STP 22.4 litres of \( CO_2 \) are produced using \( CaCO_3 \) = 100g
So, \( \frac{2 \times 273}{300} \) litres are produced by = 100/22.4 × 2274/300 = 125g
(b) 22.4 litres are \( CO_2 \) are prepared from acid = 73g
\( \frac{2 \times 273}{300} \) litres are prepared from = 73/22.4 × 2273/300 = 5.9g
In simple words: We first convert the gas volume to standard conditions, then use the chemical equation to find how much calcium carbonate and acid we need.

📝 Teacher's Note: This combines gas laws with stoichiometry. Teach students to always convert to STP first before using molar volume relationships.

🎯 Exam Tip: For gas problems, always check if conditions are at STP. If not, use gas laws to convert first, then apply stoichiometry.

Solution 7.
Answer: \( 2H_2O \rightarrow 2H_2 + O_2 \)
2 V 2 V 1 V
2 moles of \( H_2O \) gives = 1 mole of \( O_2 \)
So, 1 mole of \( H_2O \) will give = 0.5 moles of \( O_2 \)
so, mass of \( O_2 \) = no. of moles × molecular mass
= 0.5 × 32 = 16 g of \( O_2 \)
and 1 mole of \( O_2 \) occupies volume = 22.4 litre
so, 0.5 moles will occupy = 22.4 × 0.5 = 11.2 litres at S.T.P.
In simple words: When we break down water using electricity, we get hydrogen and oxygen gases. From one mole of water, we get half a mole of oxygen gas.

📝 Teacher's Note: Use this to reinforce the concept of molar ratios and how to convert between moles, mass, and volume for gases.

🎯 Exam Tip: Remember that the molar ratio comes directly from the balanced equation coefficients. Always check your ratios before calculating.

Solution 8.
Answer: \( 2Na_2O_2 + 2H_2O \rightarrow 4NaOH + O_2 \)
2 V 4 V 1 V
(a) Mol. Mass of \( Na_2O_2 \) = 2 × 23 + 2 × 16 = 78 g
Mass of 2 \( Na_2O_2 \) = 156 g
156 g \( Na_2O_2 \) gives = 160 g of NaOH (4 × 40 g)
So, 1.56 \( Na_2O_2 \) will give = 160 × 1.56/156 = 1.6 g
(b) 156 g \( Na_2O_2 \) gives = 22.4 litres of oxygen
So, 1.56 g will give = 22.4 × 1.56/156 = 0.224 litres
= 224 \( cm^3 \)
(c) 156 g \( Na_2O_2 \) gives = 32 g \( O_2 \)
So, 1.56 g \( Na_2O_2 \) will give = 32 × 1.56/156
= 32/100 = 0.32 g
In simple words: Sodium peroxide reacts with water to produce sodium hydroxide and oxygen gas. We can calculate exactly how much of each product forms from a given amount of reactant.

📝 Teacher's Note: This problem shows multiple product calculations from one reaction. Emphasize that each calculation uses the same molar ratio but different molecular masses.

🎯 Exam Tip: When multiple products are asked, set up separate proportions for each product using the same balanced equation. Don't mix up the molecular masses.

Solution 9.
Answer: \( 2NH_4Cl + Ca(OH)_2 \rightarrow CaCl_2 + 2H_2O + 2NH_3 \)
2 V 1 V 1 V 2 V
Mol. Mass of 2 \( NH_4Cl \) = 2[14 + (1 × 4) + 35.5] = 2[53.5] = 107 g
(a) 107 g \( NH_4Cl \) gives = 34 g \( NH_3 \)
So, 21.4 g \( NH_4Cl \) will give = 21.4 × 34/107 = 6.8 g \( NH_3 \)
(b) The volume of 17 g \( NH_3 \) is 22.4 litre
So, volume of 6.8 g will be = 6.8 × 22.4/17 = 8.96 litre
In simple words: When ammonium chloride reacts with lime water, it produces ammonia gas. We can calculate both the mass and volume of ammonia produced.

📝 Teacher's Note: This is a good example of a preparation reaction. Show students how to calculate both mass and volume of gaseous products.

🎯 Exam Tip: For gas calculations, remember that 17g of NH₃ = 1 mole = 22.4L at STP. Use this as a conversion factor.

Solution 10.
Answer: \( Al_4C_3 + 12 H_2O \rightarrow 3CH_4 + 4Al(OH)_3 \)
1 V 3 V 4 V
144g 3 × 22.4 l volume
Now, since 144 g of \( Al_4C_3 \) gives = 3 × 22.4 litre of \( CH_4 \)
So, 14.4 g of \( Al_4C_3 \) will give = 3 × 22.4 × 14.4 /144 = 6.72 litres \( CH_4 \)
In simple words: Aluminum carbide reacts with water to produce methane gas. We can calculate the volume of methane gas produced from a given amount of aluminum carbide.

📝 Teacher's Note: Point out that this is a hydrolysis reaction. The molar ratio is key - one mole of carbide gives three moles of methane gas.

🎯 Exam Tip: When dealing with carbides, remember they typically produce hydrocarbons when treated with water. Check the molar ratios carefully.

Solution 11.
Answer: \( MnO_2 + 4HCl \rightarrow MnCl_2 + 2H_2O + Cl_2 \)
1 V 4 V 1 V 1 V
(a) 1 mole of \( MnO_2 \) weighs = 87 g (mol. Mass)
So, 0.02 mole will weigh = 87 × 0.02 = 1.74 g \( MnO_2 \)
(b) 1 mole \( MnO_2 \) gives = 1 mole of \( MnCl_2 \)
So, 0.02 mole \( MnO_2 \) will give = 0.02 mole of \( MnCl_2 \)
(c) 1 mole \( MnCl_2 \) weighs = 126 g(mol mass)
So, 0.02 mole \( MnCl_2 \) will weigh = 126 × 0.02 g = 2.52 g
(d) 0.02 mole \( MnO_2 \) will form = 0.02 mole of \( Cl_2 \)
(e) 1 mole of \( Cl_2 \) weighs = 35.5 g
So, 0.02 mole will weigh = 71 × 0.02 = 1.42 g of \( Cl_2 \)
(f) 1 mole of chlorine gas has volume = 22.4 litres
So, 0.02 mole will have volume = 22.4 × 0.02 = 0.448 litre
(g) 1 mole \( MnO_2 \) requires HCl = 4 mole
So, 0.02 mole \( MnO_2 \) will require = 4 × 0.02 = 0.08 mole
(h) For 1 mole \( MnO_2 \), acid required = 4 mole of HCl
So, for 0.02 mole, acid required = 4 × 0.02 = 0.08 mole
Mass of HCl = 0.08 x 36.5 = 2.92 g
In simple words: This reaction shows how manganese dioxide reacts with hydrochloric acid to produce chlorine gas. All the quantities are related through the molar ratios from the balanced equation.

📝 Teacher's Note: This is an excellent comprehensive problem covering all aspects of stoichiometry. Use it to show students how one reaction can generate multiple calculation questions.

🎯 Exam Tip: For complex multi-part questions, always refer back to the balanced equation and use the 0.02 mole as your starting point for all calculations.

Solution 12.
Answer: \( N_2 + 3H_2 \rightarrow 2NH_3 \)
28g 6g 34g
28g of nitrogen requires hydrogen = 6g
2000g of nitrogen requires hydrogen = 6/28 × 2000 = 3000/7g
So mass of hydrogen left unreacted = 1000-3000/7 = 571.4g of \( H_2 \)
(b) 28g of nitrogen forms \( NH_3 \) = 34g
2000g of \( N_2 \) forms \( NH_3 \)
= 34/28 × 2000
= 2428.6g
In simple words: When nitrogen and hydrogen react to form ammonia, we can calculate how much hydrogen remains unreacted and how much ammonia is produced using the molar ratios.

📝 Teacher's Note: This is a limiting reagent problem. Show students how to identify which reactant is in excess and how much remains unreacted.

🎯 Exam Tip: For limiting reagent problems, first determine which reactant is completely consumed, then calculate the excess remaining and products formed.

Miscellaneous Exercise

Solution 1.
Answer: From equation: \( 2H_2 + O_2 \rightarrow 2H_2O \)
1 mole of Oxygen gives = 2 moles of steam
so, 0.5 mole oxygen will give = 2 × 0.5 = 1mole of steam
In simple words: When hydrogen and oxygen react to form water, one mole of oxygen produces two moles of water vapor. So half a mole of oxygen gives one mole of steam.

📝 Teacher's Note: This is a straightforward molar ratio problem. Emphasize that the coefficients in the balanced equation give us the molar ratios directly.

🎯 Exam Tip: Always write the balanced equation first, then use the coefficients to set up your molar ratios.

Solution 2.
Answer: \( 3Cu + 8HNO_3 \rightarrow 3Cu(NO_3)_2 + 4H_2O + 2NO \)
1 V 8 V 3 V 2 V
Mol. Mass of 8 \( HNO_3 \) = 8 × 63 = 504 g
(a) For 504 g \( HNO_3 \), Cu required is = 192 g
So, for 63g \( HNO_3 \) Cu required = 192 × 63/504 = 24g
(b) 504 g of \( HNO_3 \) gives = 2 × 22.4 litre volume of NO
So, 63g of \( HNO_3 \) gives = 2 × 22.4 × 63/504 = 5.6 litre of NO
In simple words: When copper reacts with nitric acid, it produces copper nitrate, water, and nitric oxide gas. We can calculate how much copper is needed and how much gas is produced.

📝 Teacher's Note: This reaction is important for understanding how metals react with acids. Point out that this produces a gas and involves redox chemistry.

🎯 Exam Tip: For acid-metal reactions, pay attention to the gas produced and its volume at STP. The molar ratios are crucial for correct calculations.

Solution 3.
Answer: (a) 28g of nitrogen = 1mole
So, 7g of nitrogen = 1/28 × 7 = 0.25 moless
(b) Volume of 71 g of \( Cl_2 \) at STP = 22.4 litres
Volume of 7.1 g chlorine = 22.4 × 7.1/71 = 2.24 litre
(c) 22400 \( cm^3 \) volume have mass = 28 g of CO(molar mass)
So, 56 \( cm^3 \) volume will have mass = 28 × 56/22400 = 0.07 g
In simple words: These calculations show how to convert between mass, moles, and volume for different gases using their molar masses and the standard molar volume.

📝 Teacher's Note: This covers all three basic conversions in gas chemistry - mass to moles, mass to volume, and volume to mass. Practice these conversion factors regularly.

🎯 Exam Tip: Remember the key conversion factors: 1 mole = molecular mass in grams = 22.4L at STP. Use these to solve any gas problem.

Solution 4.
Answer: % of N in \( NaNO_3 \) = \( \frac{14}{85} \times 100 = 16.47\% \)
% of N in \( (NH_4)_2SO_4 \) = \( \frac{14}{132} \times 100 = 21.21\% \)
% of N in \( CO(NH_2)_2 \) = \( \frac{14}{60} \times 100 = 46.66\% \)
So, highest percentage of N is in urea.
In simple words: To find the best nitrogen fertilizer, we calculate what percentage of each compound is actually nitrogen. Urea has the highest nitrogen content per gram.

📝 Teacher's Note: This is a practical application of percentage composition. Relate it to real-world fertilizer choices in agriculture.

🎯 Exam Tip: For percentage composition, the formula is (mass of element/molecular mass) × 100. Always check which compound gives the highest percentage.

Solution 5.
Answer: \( 2H_2O \rightarrow 2H_2 + O_2 \)
2 V 2 V 1 V
(a) From equation, 2 V of water gives 2 V of \( H_2 \) and 1 V of \( O_2 \)
where 2 V = 2500 \( cm^3 \)
so, volume of \( O_2 \) liberated = 2V/V = 1250 \( cm^3 \)
(b)
\( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \)
\( \frac{P_1V_1}{T_1} = \frac{7P_1 \times V_2}{2 \times T_1} \)
\( V_2 = \frac{2500 \times 2}{7} \)
\( V_2 = \frac{5000}{7} cm^3 \)
(c)
\( \frac{V_1}{V_2} = \frac{T_1}{T_2} \)
\( \frac{5000}{7 \times 2500} = \frac{T_1}{T_2} \)
\( T_2 = 3.5 T_1 \)
i.e. temperature should be increased by 3.5 times.
In simple words: When water breaks down, it produces hydrogen and oxygen gases in a 2:1 ratio. We can use gas laws to find how pressure and temperature changes affect the volume.

📝 Teacher's Note: This combines chemical stoichiometry with gas laws. Show students how to identify which gas law to apply based on what variables change.

🎯 Exam Tip: For combined gas law problems, identify which variables are constant and which change. Use the appropriate gas law equation.

Solution 6.
Answer: Molecular mass of urea = 12 + 16 + 2(14+2) = 60g
60g of urea contains nitrogen = 28g
So, in 50g of urea, nitrogen present = 23.33 g
50 kg of urea contains nitrogen = 23.33kg
In simple words: Urea contains two nitrogen atoms, so we can calculate exactly how much nitrogen is present in any given amount of urea fertilizer.

📝 Teacher's Note: This is a practical application showing how to find the actual nutrient content in fertilizers. Connect this to real farming applications.

🎯 Exam Tip: For urea problems, remember it has the formula CO(NH₂)₂, which contains 2 nitrogen atoms with atomic mass 14 each, giving 28g nitrogen per 60g urea.

Solution 7.
Answer:
(a) 80% C and 20% H
So, atomic ratio of C and H are: \( C = \frac{80}{12} = 6.66 \); \( H = \frac{20}{1} = 20 \)
Simple ratio of C:H = 1 : 3
So, empirical formula is \( CH_3 \)
(b) Empirical formula mass = 12+(3×1) = 15 g
Vapour density = 15
So, the molecular mass = 15(V.D) × 2 = 30 g
Hence, n= 2 so the molecular formula is \( C_2H_6 \)
In simple words: We find the simplest ratio of carbon to hydrogen atoms, then use vapour density to find the actual molecular formula.

📝 Teacher's Note: Demonstrate the calculation step by step on the board, emphasizing that vapour density × 2 gives molecular mass. Students often forget this multiplication factor.

🎯 Exam Tip: Always show the empirical formula calculation first, then use vapour density to find molecular formula. Include all calculation steps for full marks.

Solution 8.
Answer:
22400 cm³ \( CO_2 \) has mass = 44g
so, 224 cm³ \( CO_2 \) will have mass= 0.44 g
Now since \( CO_2 \) is being formed and X is a hydrocarbon so it contains C and H.
In 0.44g \( CO_2 \), mass of carbon=0.44-0.32=0.12g=0.01g atom
So, mass of Hydrogen in X = 0.145-0.12 = 0.025g
= 0.025g atom
Now the ratio of C:H is C=1: H=2.5 or C=2 : H=5
i.e. the formula of hydrocarbon is \( C_2H_5 \)
(a) C and H
(b) Copper (II) oxide was used for reduction of the hydrocarbon.
(c)
(i) no. of moles of \( CO_2 \)= 0.44/44 = 0.01 moles
(ii) mass of C = 0.12 g
(iii) mass of H = 0.025 g
(iv) The empirical formula of X = \( C_2H_5 \)
In simple words: When we burn a hydrocarbon completely, all carbon becomes CO₂ and all hydrogen becomes H₂O, helping us find the original formula.

📝 Teacher's Note: Use a simple analogy - like breaking down a recipe to find individual ingredients. Show students how mass conservation works in combustion analysis.

🎯 Exam Tip: Remember that in combustion analysis, mass of carbon in CO₂ = (12/44) × mass of CO₂. Show this calculation clearly.

Solution 9.
Answer:
Mass of X in the given compound =24g
Mass of oxygen in the given compound =64g
So total mass of the compound =24+64=88g
% of X in the compound = 24/88 × 100 = 27.3%
% of oxygen in the compound=64/88 × 100 =72.7%

So simplest formula = \( XO_2 \)
In simple words: We calculate percentages of each element, then find the simplest whole number ratio of atoms to get the empirical formula.

📝 Teacher's Note: Emphasize that students should always check if atomic ratios can be simplified to whole numbers. Use the "divide by smallest" method consistently.

🎯 Exam Tip: Present your work in a clear table format as shown above. Examiners appreciate organized calculations and award more marks for neat presentation.

Solution 10.
Answer:
(a) V.D = \( \frac{mass\ of\ gas\ at\ STP}{mass\ of\ equal\ volume\ of\ H_2} = \frac{8.5}{5} = 17 \)
(b) Molecular mass = 17(V.D) × 2= 34g
In simple words: Vapour density compares how heavy a gas is compared to hydrogen gas, and we use this to find molecular mass.

📝 Teacher's Note: Remind students that vapour density is always measured relative to hydrogen, not air. This is a common source of confusion.

🎯 Exam Tip: Remember the formula: Molecular mass = 2 × Vapour density. This relationship is frequently tested.

Solution 11.
Answer:
(a) \( CO_2 + C \rightarrow 2CO \)
1 V 1 V 2 V
12 g of C gives = 44.8 litre volume of CO
So, 3 g of C will give = 11.2 litre of CO
(b) \( 2CO + O_2 \rightarrow 2CO_2 \)
2 V 1 V 2 V
(i) 2 V CO requires oxygen = 1 V
so, 24 cm³ CO will require = 24/2 =12 cm³
(ii) 2 × 22400 cm³ CO gives = 2 × 22400 cm³ \( CO_2 \)
so, 24cm³ CO will give = 24 cm³ \( CO_2 \)
In simple words: These are stoichiometry problems where we use balanced equations and gas volume relationships to find how much product forms.

📝 Teacher's Note: Use the concept that equal volumes of gases contain equal moles under same conditions. Draw volume ratios clearly above balanced equations.

🎯 Exam Tip: Always write the balanced equation first, then write volume ratios below each compound. This prevents calculation errors.

Solution 12.
Answer:
\( 2Ca(NO_3)_2 \rightarrow 2CaO + 4NO_2 + O_2 \)
2 V 2 V 4 V 1 V
(a) 56 g of CaO is obtained with \( NO_2 \) = 2 × 22.4 litre of \( NO_2 \)
So, 5.6g of CaO is obtained with \( NO_2 \)=2 × 22.4 × 5.6/56
= 4.48 litre
(b) 56 g of CaO is obtained by = 164 g \( Ca(NO_3)_2 \)
So, 5.6 g CaO is obtained by = 5.6 × 56/164 g \( Ca(NO_3)_2 \)
= 16.4 g of \( Ca(NO_3)_2 \) is heated.
In simple words: We use the balanced equation to find relationships between reactants and products, then scale up or down as needed.

📝 Teacher's Note: Show students how to set up proportion calculations clearly. Emphasize checking units throughout the calculation.

🎯 Exam Tip: Use unitary method for stoichiometry calculations - "if X gives Y, then Z gives...". This approach reduces errors.

Solution 13.
Answer:
(a) Number of molecules in 100cm³ of oxygen=Y
According to Avogadro's law, Equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules. Therefore, number of molecules in 100 cm³ of nitrogen under the same conditions of temperature and pressure = Y
So, number of molecules in 50 cm³ of nitrogen under the same conditions of temperature and pressure =Y/100 × 50=Y/2
(b) (i) Empirical formula is the formula which tells about the simplest ratio of combining capacity of elements present in a compound.
(ii) The empirical formula is \( CH_3 \)
(iii) The empirical formula mass for \( CH_2O \) = 30
V.D = 30
Molecular formula mass = V.D × 2 = 60
Hence, n =mol. Formula mass/empirical formula mass= 2
So, molecular formula = \( (CH_2O)_2 = C_2H_4O_2 \)
In simple words: Avogadro's law tells us that equal volumes of any gas have equal molecules, so we can easily compare different gases.

📝 Teacher's Note: Use balloons or gas syringes to demonstrate Avogadro's law practically. This helps students visualize the concept better.

🎯 Exam Tip: State Avogadro's law clearly before applying it. Examiners look for proper law statements in gas-related problems.

Solution 14.
Answer:
The relative atomic mass of Cl = (35 × 3 + 1 × 37)/4=35.5 amu
In simple words: We calculate average atomic mass by considering the abundance (how much) of each isotope and their individual masses.

📝 Teacher's Note: Explain that this is like calculating a weighted average - more abundant isotopes contribute more to the final average.

🎯 Exam Tip: The formula is: (mass₁ × abundance₁ + mass₂ × abundance₂)/total abundance. Show all steps clearly.

Solution 15.
Answer:
Mass of silicon in the given compound =5.6g
Mass of the chlorine in the given compound=21.3g
Total mass of the compound=5.6g+21.3g=26.9g
% of silicon in the compound = 56/26.9 × 100 = 20.82%
% of chlorine in the compound = 21.2/26.9 × 100 = 79.18%

So the empirical formula of the given compound =\( SiCl_3 \)
In simple words: By finding the percentage of each element and converting to atomic ratios, we determine the simplest formula of the compound.

📝 Teacher's Note: Remind students to use correct atomic masses from the periodic table. Small errors in atomic masses can affect the final answer.

🎯 Exam Tip: Round atomic ratios sensibly - 2.23 is close to 2, but 0.74 is close to 1. Practice recognizing these patterns.

Solution 16.
Answer:

So, empirical formula is \( PH_2O_3 \) or \( H_2PO_3 \)
Empirical formula mass = 31+ 2 × 1 + 3 × 16 = 81
The molecular formula is = \( H_4P_2O_6 \) because n = 162/81=2
In simple words: We find the simplest ratio of atoms, then use molecular mass to determine if we need to multiply this ratio.

📝 Teacher's Note: Show students how to convert empirical formula to molecular formula using the n-factor method. This is a key skill.

🎯 Exam Tip: Always check if n = molecular mass/empirical formula mass is a whole number. If not, recheck your calculations.

Solution 17.
Answer:
\( V_1 = 10 \) litres \( V_2 = ? \)
\( T_1= 27+ 273 = 300K \) \( T_2=273K \)
\( P_1=700 \) mm \( P_2 = 760 \) mm
Using the gas equation
\[ \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \]
\[ V_2 = \frac{P_1V_1T_2}{T_1P_2} = \frac{700 \times 10 \times 273}{300 \times 760} \]
Molecular weight A = 60
So, weight of 22.4 litres of A at STP =60 g
Weight of = \( \frac{700 \times 10 \times 273}{300 \times 760} \) litres of A at STP
= \( \frac{60 \times 700 \times 10 \times 273}{22.4 \times 300 \times 760} \) g or 22.45g
In simple words: We use the combined gas law to find the volume at different conditions, then calculate mass using molar mass relationships.

📝 Teacher's Note: Teach students to always convert temperature to Kelvin and keep pressure units consistent. These are common mistake areas.

🎯 Exam Tip: Write down all given values clearly before starting calculations. This helps avoid substitution errors in the formula.

Solution 18.
Answer:
(a) Molecular mass of \( CO_2 \) = 12+ 2×16 = 44 g
So, vapour density (V.D) = mol. Mass/2 = 44/2 = 22
V.D = \( \frac{mass\ of\ certain\ amount\ of\ CO_2}{mass\ of\ equal\ volume\ of\ hydrogen} = \frac{m}{1} \)
\[ 22 = \frac{m}{1} \]
So, mass of \( CO_2 \) = 22 kg
(b) According to Avogadro's law, equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.
So, number of molecules of carbon dioxide in the cylinder =number of molecules of hydrogen in the cylinder=X
In simple words: Vapour density helps us compare gas masses, and Avogadro's law tells us equal volumes have equal molecules.

📝 Teacher's Note: Connect vapour density to everyday examples - helium balloons float because helium has lower vapour density than air.

🎯 Exam Tip: Remember that vapour density = molecular mass/2. This is one of the most frequently tested relationships.

Solution 19.
Answer:
(a) The volume occupied by 1 mole of chlorine = 22.4 litre
(b) Since PV=constant so, if pressure is doubled; the volume will become half i.e. 11.2 litres.
(c) \( V_1/V_2 = T_1/T_2 \)
22.4/\( V_2 \) =273/546
\( V_2 \) = 44.8 litres
(d) Mass of 1 mole \( Cl_2 \) gas =35.5 × 2 =71 g
In simple words: Gas volumes change with temperature and pressure according to simple mathematical relationships that help us predict behavior.

📝 Teacher's Note: Use gas syringes or balloons to demonstrate how gas volume changes with temperature and pressure. Visual demonstrations stick better.

🎯 Exam Tip: For gas law problems, always identify which quantities are constant and which are changing. This guides which law to use.

Solution 20.
Answer:
(a) Total molar mass of hydrated \( CaSO_4 \cdot xH_2O \) = 136+18x
Since 21% is water of crystallization, so
\[ \frac{18x}{136 + 18x} = \frac{21}{100} \]
So, x = 2 i.e. water of crystallization is 2.
(b) For 18 g water, vol. of hydrogen needed = 22.4 litre
So, for 1.8 g, vol. of \( H_2 \) needed= 1.8 × 22.4/18 = 2.24 litre
Now 2 vols. of water = 1 vol. of oxygen
1 vol. of water =1/2 vol. of \( O_2 \) =22.4/2=11.2 lit.
18 g of water = 11.2 lit. of \( O_2 \)
1.8 g of water = 11.2/18 × 18/10=1.12 lit.
(c) 32g of dry oxygen at STP = 22400cc
2g will occupy = 22400/2/32=1400cc
\( P_1=760mm \) \( P_2 =740mm \)
\( V_1=1400cc \) \( V_2 =? \)
\( T_1 =273 K \), \( T_2 = 27 +73 = 300K \)
\[ \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \]
\[ V_2 = \frac{P_1V_1T_2}{T_1P_2} = \frac{760 \times 1400 \times 300}{273 \times 740} = 1580 cc \]
= 1580/1000 =1.58l
(d) \( P_1= 750mm \) \( P_2=760mm \)
\( V_1= 44lit. \) \( V_2=? \)
\( T_1= 298K \) \( T_2=273K \)
\[ \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \]
\[ V_2 = \frac{P_1V_1T_2}{T_1P_2} = \frac{750 \times 44 \times 273}{298 \times 760} = 39.78 lit. \]
22.4 lit. of CO2 at STP has mass= 44g
39.78 lit. of CO2 at STP has mass =44 × 39.78/22.4
=78.14 g
(e) Since 143.5g of AgCl is produced from =58.5 g of NaCl
so, 1.435 g of AgCl is formed by =0.585 g of NaCl
% of NaCl =0.585 ×100 = 58.5%
In simple words: This problem combines several chemistry concepts - hydration, gas laws, and precipitation reactions to solve different parts step by step.

📝 Teacher's Note: This is a multi-part problem requiring different chemistry concepts. Break it down into clear sections and solve each part separately to avoid confusion.

🎯 Exam Tip: For complex multi-part questions, clearly label each part (a), (b), (c) etc. and show all calculation steps. Partial marks are available for correct method even if final answer is wrong.

Solution 21.
Answer: \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \)
\( \frac{P_1 \times 22.4}{273} = \frac{2P_2 V_2}{546} \)
\( V_2 = 22.4 \) litre
In simple words: Using the gas law equation, when temperature doubles, the volume of gas also doubles at constant pressure.

📝 Teacher's Note: Help students visualize this with a balloon - when you heat it, the air inside expands and the balloon gets bigger. This is Charles's Law in action.

🎯 Exam Tip: Always convert temperature to Kelvin (°C + 273) before using in gas law calculations. This is a common error students make.

Solution 22.
Answer: (a) The molecular mass of \( (Mg(NO_3)_2.6H_2O \) = 256.4 g
% of Oxygen = 12 × 16/256 = 75%
(b) The molecular mass of boron in \( Na_2B_4O_7.10H_2O \) = 382 g
% of B = 4 × 11/382 = 11.5%
In simple words: To find percentage of an element, divide its total atomic mass in the compound by the molecular mass and multiply by 100.

📝 Teacher's Note: Emphasize counting all atoms of each element carefully. In part (a), there are 12 oxygen atoms total (6 from nitrate + 6 from water).

🎯 Exam Tip: Always list out all atoms of each element separately before calculating - this prevents counting errors.

Solution 23.
Answer: \( V \times \frac{760}{273} = \frac{360 \times 380}{360} \)
\( V = \frac{360 \times 380 \times 273}{760 \times 360} = 136.5cm^3 \)
136.5cm³ of the gas weighs = 0.546
22400 cm³ of the gas weight = \( \frac{0.546 \times 22400}{136.5} = 89.6 \) a.m.u
Relative molecular mass = 89.6 a.m.u
In simple words: We use the gas equation to find volume at standard conditions, then calculate how much 22400 cm³ would weigh to get molecular mass.

📝 Teacher's Note: This combines gas laws with molar mass calculation. Show students that 22.4L = 22400cm³ is the molar volume at STP.

🎯 Exam Tip: Remember that relative molecular mass equals the mass of 22400 cm³ of gas at STP. Keep units consistent throughout.

Solution 24.
Answer: (a) 252 g of solid ammonium dichromate decomposes to give 152 g of solid chromium oxide, so the loss in mass in terms of solid formed = 100 g
Now, if 63 g ammonium dichromate is decomposed, the loss in mass would be = 100 × 63/252 = 25 g
(b) If 252 g of ammonium dichromate produces \( Cr_2O_3 \) = 152 g
So, 63 g ammonium dichromate will produce = 63 × 152/252 = 38 g
In simple words: Using simple proportion - if a certain amount of reactant gives a specific product, we can find what a smaller amount will give.

📝 Teacher's Note: This is stoichiometry in action. Help students see the pattern: set up ratios and cross-multiply to solve.

🎯 Exam Tip: Always write the given ratio first, then set up your unknown in the same format. This prevents confusion in calculations.

Solution 25.
Answer: \( 2H_2S + 3O_2 \rightarrow 2H_2O + 2SO_2 \)
2V 3V 2V
128 g of \( SO_2 \) gives = 2 × 22.4 litres volume
So, 12.8 g of \( SO_2 \) gives = 2 × 22.4 × 12.8/128 = 4.48 litre volume
Or one can say 4.48 litres of hydrogen sulphide.
2 × 22.4 litre \( H_2S \) requires oxygen = 3 × 22.4 litre
So, 4.48 litres \( H_2S \) will require = 6.72 litre of oxygen
In simple words: From the balanced equation, we can find how much of each gas is needed or produced using volume ratios.

📝 Teacher's Note: Show students how coefficients in balanced equations give volume ratios directly for gases at same conditions.

🎯 Exam Tip: Always balance the chemical equation first, then use coefficients as volume ratios for gas calculations.

Solution 26.
Answer: From equation, \( 2NH_3 + 2O_2 \rightarrow 2NO + 3H_2O \)
When 60 g NO is formed, mass of steam produced = 54 g
So, 1.5 g NO is formed, mass of steam produced = 54 × 1.5/60 = 1.35 g
In simple words: Using the balanced equation, we find the mass ratio between products and apply it to find the unknown mass.

📝 Teacher's Note: This shows how to use stoichiometry with actual masses rather than just moles. Emphasize the importance of the balanced equation.

🎯 Exam Tip: When given masses, convert using the molar mass ratio from the balanced equation. Set up proportions carefully.

Solution 27.
Answer: In 1 hectare of soil, \( N_2 \) removed = 20 kg
So, in 10 hectare \( N_2 \) removed = 200 kg
The molecular mass of \( Ca(NO_3)_2 \) = 164
Now, 28 g \( N_2 \) present in fertilizer = 164 g \( Ca(NO_3)_2 \)
So, 200000 g of \( N_2 \) is present in = 164 × 200000/28 = 1171.42 kg
In simple words: We calculate how much fertilizer contains the required amount of nitrogen by using the proportion of nitrogen in the compound.

📝 Teacher's Note: This connects chemistry to real agriculture. Show students how farmers actually use this math to decide fertilizer amounts.

🎯 Exam Tip: Remember that \( Ca(NO_3)_2 \) contains 2 atoms of nitrogen, so 2 × 14 = 28 g nitrogen per molecule.

Solution 28.
Answer: (a) 1 mole of phosphorus atom = 31 g of phosphorus
31 g of P = 1 mole of P
6.2g of P = \( \frac{6.2 \times 1}{31} \) = 0.2 mole of P
(b) 31 g P reacts with \( HNO_3 \) = 315 g
so, 6.2 g P will react with \( HNO_3 \) = 315 × 6.2/31 = 63 g
(c) Moles of steam formed from 31g phosphorus = 18g/18g = 1mol
Moles of steam formed from 6.2 g phosphorus = 1mol/31g×6.2=0.2 mol
Volume of steam produced at STP =0.2 × 22.4 l/MOL=4.48 litre
Since the pressure (760mm) remains constant, but the temperature (273+273)=546 is double, the volume of the steam also gets doubled
So,Volume of steam produced at 760mm Hg and 273°C = 4.48 × 2 = 8.96litre
In simple words: We use moles to find how much steam forms, then apply gas laws to find volume at different temperatures.

📝 Teacher's Note: This combines stoichiometry with gas law applications. Break it into steps: moles → volume at STP → volume at new temperature.

🎯 Exam Tip: When temperature doubles (in Kelvin), volume doubles at constant pressure. Always convert °C to Kelvin first.

Solution 29.
Answer: (a) 1 mole of gas occupies volume = 22.4 litre
(b) 112cm³ of gaseous fluoride has mass = 0.63 g
so, 22400cm³ will have mass = 0.63 × 22400/112 = 126 g
The molecular mass = At mass P + At. mass of F
126 = 31 + At. Mass of F
So, At. Mass of F = 95 g
But, at. mass of F = 19 so 95/19 = 5
Hence, there are 5 atoms of F so the molecular formula = \( PF_5 \)
In simple words: We find the molecular mass from gas density, then figure out how many fluorine atoms are present to get the formula.

📝 Teacher's Note: This shows how gas density can reveal molecular formulas. Emphasize that 22.4L is the molar volume for any gas at STP.

🎯 Exam Tip: Molecular mass = mass of 22.4L of gas at STP. Use this to find unknown atoms in molecular formulas.

Solution 30.
Answer: \( Na_2CO_3.10H_2O \rightarrow Na_2CO_3 + 10H_2O \)
286 g 106 g
So, for 57.2 g \( Na_2CO_3.10H_2O \) = 106 × 57.2/286 = 21.2 g \( Na_2CO_3 \)
In simple words: When washing soda loses its water molecules on heating, we can calculate how much anhydrous salt remains using simple proportion.

📝 Teacher's Note: This shows hydrated salts losing water. Have students observe this practically with copper sulfate crystals changing color.

🎯 Exam Tip: In dehydration reactions, focus on the mass ratio between hydrated and anhydrous forms from the balanced equation.

Solution 31.
Answer: (a) The molecular mass of \( Ca(H_2PO_4)_2 \) = 234
The % of P = 2 × 31/234 = 26.49 %
(b) Simple ratio of M = 34.5/56 = 0.616 = 1
Simple ratio of Cl = 65.5/35.5 = 1.845 = 3
Empirical formula = \( MCl_3 \)
Empirical formula mass = 162.5, Molecular mass = 2 × V.D = 325
So, n = 2
So, molecular formula = \( M_2Cl_6 \)
In simple words: We find the simplest ratio of atoms, then use molecular mass to get the actual molecular formula.

📝 Teacher's Note: Show students how to find the simplest whole number ratios by dividing by the smallest ratio. This is key to empirical formulas.

🎯 Exam Tip: Always check if your empirical formula mass divides evenly into the molecular mass to find the multiplier (n).

Solution 32.
Answer: \( V_1/V_2 = n_1/n_2 \)
So, no. of moles of Cl = x/2 (since V is directly proportional to n)
No. of moles of \( NH_3 \) = x
No. of moles of \( SO_2 \) = x/4
This is because of Avogadros law which states Equal volumes of all gases, under similar conditions of temperature and pressure, contain equal number of molecules.
So, 20 litre nitrogen contains x molecules
So, 10 litre of chlorine will contain = x × 10/20=x/2 mols.
And 20 litre of ammonia will also contain =x molecules
And 5 litre of sulphur dioxide will contain = x × 5/20 = x/4 mols.
In simple words: Equal volumes of gases contain equal numbers of molecules, so we can find mole ratios from volume ratios.

📝 Teacher's Note: This demonstrates Avogadro's law practically. Use balloons of equal size to show equal volumes containing equal molecules.

🎯 Exam Tip: Remember Avogadro's law: equal volumes = equal moles for gases at same temperature and pressure.

Solution 33.
Answer: \( 4N_2O + CH_4 \rightarrow CO_2 + 2H_2O + 4N_2 \)
4V 1V 1V 2V 4V
2 × 22400 litre steam is produced by \( N_2O \) = 4 × 22400 cm³
So, 150 cm³ steam will be produced by= 4 × 22400 × 150/2 × 22400 = 300 cm³ \( N_2O \)
In simple words: Using the balanced equation, we find the volume ratio between steam and nitrous oxide to calculate the required volume.

📝 Teacher's Note: This shows how balanced equations give direct volume relationships for gases. Emphasize using coefficients as volume ratios.

🎯 Exam Tip: The coefficients in balanced equations directly give volume ratios for gases at the same conditions.

Solution 34.
Answer: (a) Volume of \( O_2 \) = V
Since \( O_2 \)and \( N_2 \) have same no. of molecules = x
so, the volume of \( N_2 \)= V
(b) 3x molecules means 3V volume of CO
(c) 32 g oxygen is contained in = 44 g of \( CO_2 \)
So, 8 g oxygen is contained in = 44 x 8/32 = 11 g
(d) Avogadro's law is used in the above questions.
In simple words: Equal numbers of gas molecules occupy equal volumes, and we can use this to solve various gas problems.

📝 Teacher's Note: This reinforces Avogadro's law through multiple applications. Show how it connects molecular counts to volumes and masses.

🎯 Exam Tip: When questions mention "same number of molecules," immediately think of Avogadro's law and equal volumes.

Solution 35.
Answer: (a) 444 g is the molecular formula of \( (NH_4)_2PtCl_6 \)
% of Pt = (195/444) x 100 = 43.91% or 44%
(b) simple ratio of Na = 42.1/23 = 1.83 = 3
simple ratio of P = 18.9/31 = 0.609 = 1
simple ratio of O = 39/16 = 2.43 = 4
So, the empirical formula is \( Na_3PO_4 \)
In simple words: We calculate percentage by mass and find empirical formulas by converting percentages to simple whole number ratios.

📝 Teacher's Note: Show students how to round ratios to nearest whole numbers when they're close (like 1.83 ≈ 2, but here it's actually 3 when properly calculated).

🎯 Exam Tip: Always divide all ratios by the smallest one to get the simplest whole number ratios for empirical formulas.

Solution 36.
Answer: \( CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \)
1V 2V 1V 2V
From equation:
22.4 litres of methane requires oxygen = 44.8 litres \( O_2 \)
\( 2H_2 + O_2 \rightarrow 2H_2O \)
2V 1V 2V
From equation,
44.8 litres hydrogen requires oxygen = 22.4 litres \( O_2 \)
So, 11.2 litres will require = 22.4 × 11.2/44.8 = 5.6 litres
Total volume = 44.8 + 5.6 = 50.4 litres
In simple words: We calculate oxygen needed for both reactions separately, then add them together to get the total oxygen required.

📝 Teacher's Note: This involves multiple reactions. Teach students to handle each reaction separately, then combine results systematically.

🎯 Exam Tip: For multiple reactions, solve each one step-by-step and clearly label which calculation belongs to which reaction.

Solution 37.
Answer: According to Avogadro's law: Equal volumes of all gases, under similar conditions of temperature and pressure, contain equal number of molecules.

So, 1 mole of each gas contains = \( 6.02 \times 10^{23} \) molecules

Mol. Mass of \( H_2 \)(2), \( O_2 \)(32), \( CO_2 \)(44), \( SO_2 \)(64), \( Cl_2 \)(71)

(1) Now 2 g of hydrogen contains molecules = \( 6.02 \times 10^{23} \)

So, 8g of hydrogen contains molecules = \( 8/2 \times 6.02 \times 10^{23} \)

= \( 4 \times 6.02 \times 10^{23} \) = 4M molecules

(2) 32g of oxygen contains molecules = \( 8/32 \times 6.02 \times 10^{23} \) = M/4

(3) 44g of carbon dioxide contains molecules = \( 8/44 \times 6.02 \times 10^{23} \) = 2M/11

(4) 64g of sulphur dioxide contains molecules = \( 6.02 \times 10^{23} \)

So, 8g of sulphur dioxide molecules = \( 8/64 \times 6.02 \times 10^{23} \) = M/8

(5) 71 g of chlorine contains molecules = \( 6.02 \times 10^{23} \)

So, 8g of chlorine molecules = \( 8/72 \times 6.02 \times 10^{23} \) = 8M/71

Since 8M/71 < M/8 < 2M/11 < M/4 < 4M

Thus \( Cl_2 < SO_2 < CO_2 < O_2 < H_2 \)

(i) Least number of molecules in \( Cl_2 \)

(ii) Most number of molecules in \( H_2 \)
In simple words: When comparing equal masses of different gases, lighter gases have more molecules because each molecule weighs less, so you can fit more of them into the same mass.

📝 Teacher's Note: Use the analogy of coins - comparing 8 grams of pennies vs 8 grams of quarters. Since pennies are lighter, you get more pennies in 8 grams. Students often forget to divide mass by molecular mass first.

🎯 Exam Tip: Always write "Number of molecules = (given mass/molar mass) × 6.02 × 10²³" as your first step - this formula gets you marks even if calculations go wrong.

Solution 38.
Answer: \( Na_2SO_4 + BaCl_2 \rightarrow BaSO_4 + 2NaCl \)

Molecular mass of \( BaSO_4 \) = 233 g

Now, 233 g of \( BaSO_4 \) is produced by \( Na_2SO_4 \) = 142 g

So, 6.99 g \( BaSO_4 \) will be produced by = \( 6.99 \times 142/233 \) = 4.26

The percentage of \( Na_2SO_4 \) in original mixture = \( 4.26 \times 100/10 \)

= 42.6%
In simple words: We use the chemical equation to find how much sodium sulphate was needed to make the observed amount of barium sulphate precipitate, then calculate what percentage this represents of the original mixture.

📝 Teacher's Note: Emphasize that stoichiometry problems always start with a balanced equation. Show students how to set up the proportion using molar masses from the balanced equation.

🎯 Exam Tip: Write the balanced equation first, then use the ratio: (mass of product formed) × (molar mass of reactant)/(molar mass of product) = mass of reactant needed.

Solution 39.
Answer: (a) 1 litre of oxygen has mass = 1.32 g
So, 24 litres (molar vol. at room temp.) will have mass = \( 1.32 \times 24 \)

= 31.6 or 32 g

(b) \( 2KMnO_4 \rightarrow K_2MnO_4 + MnO_2 + O_2 \)

316 g of \( KMnO_4 \) gives oxygen = 24 litres

So, 15.8 g of \( KMnO_4 \) will give = \( 24 \times 316/15.8 \) = 1.2 litres
In simple words: Part (a) finds the molar mass of oxygen gas at room temperature. Part (b) uses stoichiometry to calculate how much oxygen gas is produced when potassium permanganate decomposes.

📝 Teacher's Note: Remind students that molar volume at room temperature (24 L) is different from STP (22.4 L). The decomposition of KMnO₄ is a classic example - use it to reinforce stoichiometric calculations.

🎯 Exam Tip: Always check if the question mentions STP or room temperature - the molar volumes are different (22.4 L vs 24 L respectively).

Solution 40.
Answer: (a)
(i) The no. of moles of \( SO_2 \) = \( 3.2/64 \) = 0.05 moles
(ii) In 1 mole of \( SO_2 \), no. of molecules present = \( 6.02 \times 10^{23} \)
So, in 0.05 moles, no. of molecules = \( 6.02 \times 10^{23} \times 0.05 \)

= \( 3.0 \times 10^{22} \)

(iii) The volume occupied by 64 g of \( SO_2 \) = \( 22.4 \, dm^3 \)

\( 3.2 \, g \) of \( SO_2 \) will be occupied by volume = \( 22.4 \times 3.2/64 \) = \( 1.12 \, dm^3 \)

(b) Gram atoms of Pb = \( 6.21/207 \) = 0.03 = 1

Gram atoms of Cl = \( 4.26/35.5 \) = 0.12 = 4

So, the empirical formula = \( PbCl_4 \)
In simple words: This problem shows how to calculate moles, molecules, and volume from a given mass of gas, plus how to find the simplest formula of a compound by comparing the ratios of atoms.

📝 Teacher's Note: Break this into clear steps: mass → moles → molecules → volume. For empirical formula, show how dividing by atomic masses gives atom ratios, then simplify to whole numbers.

🎯 Exam Tip: For empirical formula questions, always divide mass by atomic mass first, then find the simplest whole number ratio by dividing by the smallest value.

Solution 41.
Answer: (i) D contains the maximum number of molecules because volume is directly proportional to the number of molecules.

(ii) The volume will become double because volume is directly proportional to the no. of molecules at constant temperature and pressure.
\( V_1/V_2 = n_1/n_2 \)
\( V_1/V_2 = n_1/2n_1 \)
So, \( V_2 = 2V_1 \)

(iii) Gay Lussac's law of combining volume is being observed.

(iv) The volume of D = \( 5.6 \times 4 \) = \( 22.4 \, dm^3 \), so the number of molecules = \( 6 \times 10^{23} \) because according to mole concept 22.4 litre volume at STP has = \( 6 \times 10^{23} \) molecules

(v) No. of moles of D = 1 because volume is 22.4 litre
so, mass of \( N_2O \) = \( 1 \times 44 \) = 44 g
In simple words: This question applies Avogadro's law - equal volumes of gases contain equal numbers of molecules at same temperature and pressure. Container D has the largest volume, so it has the most molecules.

📝 Teacher's Note: Use this problem to reinforce Avogadro's law and gas relationships. Students should understand that volume ratio = molecule ratio at constant T and P.

🎯 Exam Tip: Remember that 22.4 L at STP always contains 6.02 × 10²³ molecules (1 mole) for any gas - this is a key relationship that appears in many gas problems.

Solution 42.
Answer: (a) \( NaCl + NH_3 + CO_2 + H_2O \rightarrow NaHCO_3 + NH_4Cl \)
\( 2NaHCO_3 \rightarrow Na_2CO_3 + H_2O + CO_2 \)

From equation:

106 g of \( Na_2CO_3 \) is produced by = 168 g of \( NaHCO_3 \)

So, 21.2 g of \( Na_2CO_3 \) will be produced by = \( 168 \times 21.2/106 \)

= 33.6 g of \( NaHCO_3 \)

(b) For 84 g of \( NaHCO_3 \), required volume of \( CO_2 \) = 22.4 litre

So, for 33.6 g of \( NaHCO_3 \), required volume of \( CO_2 \) = \( 22.4 \times 33.6/84 \)

= 8.96 litre
In simple words: This shows the Solvay process where sodium carbonate is made from sodium hydrogen carbonate. We calculate backwards from the final product to find how much starting material and carbon dioxide we need.

📝 Teacher's Note: Connect this to the industrial Solvay process for making washing soda. Emphasize how stoichiometry helps calculate industrial quantities and costs.

🎯 Exam Tip: In two-step reactions, always identify which equation gives you the relationship you need. Here, the second equation relates NaHCO₃ to Na₂CO₃.

Solution 43.
Answer: (a) \( NH_4NO_3 \rightarrow N_2O + 2H_2O \)
1 mole 1 mole 2 mole
1 V 1 V 2 V
44.8 litres of water produced by = 22.4 litres of \( NH_4NO_3 \)
So, 8.96 litres will be produced by = \( 22.4 \times 8.96/44.8 \)
= 4.48 litres of \( NH_4NO_3 \)
So, 4.48 litres of \( N_2O \) is produced.
(i) 44.8 litre \( H_2O \) is produced by = 80 g of \( NH_4NO_3 \)
So, 8.96 litre \( H_2O \) will be produced by = \( 80 \times 8.96/44.8 \)
= 16g \( NH_4NO_3 \)
(iii) % of O in \( NH_4NO_3 \) = \( 3 \times 16/80 \) = 60%
In simple words: When ammonium nitrate decomposes, it produces nitrous oxide and water vapour. We use the balanced equation to calculate how much of each product forms.

📝 Teacher's Note: This decomposition reaction is important in both laboratory and industrial contexts. Show students how volume ratios match mole ratios for gases.

🎯 Exam Tip: For gas reactions, remember that volume ratios = mole ratios when temperature and pressure are constant. This makes calculations much easier.

Solution 44.
Answer: (a)

so, empirical formula is \( K_2BeF_4 \)

(b) \( 3CuO + 2NH_3 \rightarrow 3Cu + 3H_2O + N_2 \)
3 V 2 V 3 V 1V
\( 3 \times 80 \) g of CuO reacts with = \( 2 \times 22.4 \) litre of \( NH_3 \)
so, 120 g of CuO will react with = \( 2 \times 22.4 \times 120/(80 \times 3) \)
= 22.4 litres
In simple words: Part (a) finds the simplest formula by converting percentages to atom ratios. Part (b) calculates how much ammonia gas is needed to reduce a given amount of copper oxide.

📝 Teacher's Note: For empirical formula, emphasize dividing by atomic mass, then by the smallest value. For gas stoichiometry, show how mole ratios from equations become volume ratios.

🎯 Exam Tip: In empirical formula problems, always make a table with columns for element, %, atomic mass, atomic ratio, and simple ratio - this organized approach prevents errors.

Solution 45.
Answer: (a) The molecular mass of ethylene (\( C_2H_4 \)) is 28 g
No. of moles = \( 1.4/28 \) = 0.05 moles
No. of molecules = \( 6.023 \times 10^{23} \times 0.05 \) = \( 3 \times 10^{22} \) molecules
Volume = \( 22.4 \times 0.05 \) = 1.12 litres

(b) Molecular mass = \( 2 \times V.D \)
So, V.D = \( 28/2 \) = 14
In simple words: Given the mass of ethylene gas, we calculate how many molecules it contains and what volume it occupies at standard conditions. We also find its vapour density.

📝 Teacher's Note: This is a comprehensive mole concept problem covering mass-mole-molecule-volume relationships. Remind students that vapour density = molecular mass/2.

🎯 Exam Tip: Always follow the sequence: given mass → moles → molecules AND moles → volume. Remember vapour density = molecular mass ÷ 2.

Solution 46.
Answer: (a) Molecular mass of \( Na_3AlF_6 \) = 210
So, Percentage of Na = \( 3 \times 23 \times 100/210 \) = 32.85%
(b) \( 2CO + O_2 \rightarrow 2CO_2 \)
2 V 1 V 2 V
1 mole of \( O_2 \) has volume = 22400 ml
Volume of oxygen used by 2 × 22400 ml CO = 22400 ml
So, Vol. of \( O_2 \) used by 560 ml CO = \( 22400 \times 560/(2 \times 22400) \)

= 280 ml

So, Volume of \( CO_2 \) formed is 560 ml.
In simple words: Part (a) calculates what percentage of cryolite's mass comes from sodium atoms. Part (b) uses Gay-Lussac's law to find gas volumes in the combustion of carbon monoxide.

📝 Teacher's Note: Connect cryolite to aluminum production - it's used as a flux. For gas volumes, emphasize that volume ratios equal coefficient ratios in balanced equations.

🎯 Exam Tip: For percentage composition, use: (number of atoms × atomic mass × 100) ÷ molecular mass. For gas volume calculations, volume ratios = mole ratios from equation.

Solution 47.
Answer: a. Mass of gas X = 10g
Mass of hydrogen gas = 2
Relative vapour density

= Mass of volume of gas X under similar conditions / Mass of volume of hydrogen gas under similar conditions = 10/2 = 5

Relative molecular mass of the gas = 2 × relative vapour density = 2 × 5

= 10
b.
i. The combustion reaction \( 2C_2H_2(g) + 5O_2(g) \rightarrow 4CO_2(g) + 2H_2O(g) \)
According to Gay-Lussac's law,
2 volume of acetylene requires 5 volume of oxygen to burn it
∴ 1 volume of acetylene requires 2.5 volume of oxygen to burn it
∴ \( 200cm^3 \) requires \( 2.5 \times 200 = 500 cm^3 \) of oxygen
2 volume of acetylene on combustion gives \( 4CO_2 \)
∴ 1 volume of acetylene on combustion gives \( 2CO_2 \)
∴ 200cc of acetylene on combustion will give \( 200 \times 2 = 400cc \) of \( CO_2 \)
ii. Hydrogen = 12.5%
∴ Nitrogen = 100-12.5 = 87.5%

The Empirical formula of the compound is \( NH_2 \)
Empirical formula weight = 14+2 = 16
Relative molecular mass = 37
In simple words: This problem covers vapour density calculations, combustion analysis of acetylene, and empirical formula determination from percentage composition data.

📝 Teacher's Note: Vapour density problems confuse students - emphasize that it's just comparing gas masses at same conditions. For combustion, show how volume ratios come directly from balanced equations.

🎯 Exam Tip: Vapour density = molecular mass ÷ 2. For empirical formula from %, convert to mass ratios, then atomic ratios, then simplest whole number ratios.

Solution 48.
Answer: N = \( \frac{\text{Relative molecular mass}}{\text{Empirical Weight}} = \frac{37}{16} = 2.3 \approx 2 \)
Molecular formula = n × empirical formula = 2 × NH₂ = N₂H₄

c.
i. Molecules of nitrogen gas in a cylinder = 24 × 10²⁴
Avogadro's number = 6 × 10²³

1. Mass of nitrogen in a cylinder = \( \frac{24 \times 10^{24} \times 28}{6 \times 10^{23}} \) = 1120g

2. Volume of nitrogen at stp
Volume of 28 g of N₂ = 22.4dm³

Volume of 1120g of N₂ = \( \frac{1120 \times 22.4}{28} \) = 896 dm³

📝 Teacher's Note: Help students remember that Avogadro's number (6 × 10²³) is the bridge between molecular count and moles. Use dimensional analysis to show how units cancel properly in these calculations.

🎯 Exam Tip: Always write the units clearly in each step and check that your final answer makes sense - 1120g of nitrogen should indeed occupy a large volume at STP.

Solution 48 continued.
Answer: a.
i. 10 litres of LPG contains
Propane = \( \frac{60}{100} \times 10 = 6 \) litres
Butane = \( \frac{40}{100} \times 10 = 4 \) litres

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
1 vol.            3 vol.
6L                  18L

2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
2 vol.             8 vol.
4L                  16L
18+16=34 L

ii. Molecular mass of NH₄(NO₃) = 80
H=1, N=14, O=16
% of Nitrogen
As 80 g of NH₄(NO₃) contains 28 g of nitrogen
∴ 100 g of NH₄(NO₃) will contain \( \frac{28 \times 100}{80} \) = 35%

% of Oxygen
As 80 g of NH₄(NO₃) contains 48 g of oxygen
∴ 100 g of NH₄(NO₃) will contain \( \frac{100 \times 48}{80} \) = 60%

📝 Teacher's Note: Emphasize that in combustion reactions, coefficients represent volume ratios for gases. Show students how to track each element through percentage composition calculations.

🎯 Exam Tip: In percentage composition problems, always verify that your percentages add up to 100% as a quick check for calculation errors.

Solution 48 continued.
Answer: b.
i. Equation for reaction of calcium carbonate with dilute hydrochloric acid:
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂ ↑

ii. Relative molecular mass of calcium carbonate = 100
Mass of 4.5 moles of calcium carbonate
= No. of moles × Relative molecular mass
= 4.5 × 100
= 450g

iii. CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂ ↑
As, 100g of calcium carbonate gives 22.4dm³ of CO₂
∴ 450 g of calcium carbonate will give \( \frac{450 \times 22.4}{100} \) = 100.8 L

iii. Molecular mass of calcium carbonate = 100
Relative molecular mass of calcium chloride = 111
As 100 g of calcium carbonate gives 111g of calcium chloride
∴ 450 g of calcium carbonate will give \( \frac{450 \times 111}{100} \) = 499.5 g

iv.
Molecular mass of HCl = 36.5
Molecular mass of calcium carbonate = 100
As 100 g of calcium carbonate gives (2×36.5) = 73g of HCl
∴ 450 g of calcium carbonate will give \( \frac{450 \times 73}{100} \) = 328.5g

Number of moles of HCl = \( \frac{\text{Weight of HCl}}{\text{Molecular weight of HCl}} \)
= \( \frac{328.5}{36.5} \)
= 9 moles

📝 Teacher's Note: Connect stoichiometry to the balanced equation - show how mole ratios from the equation translate to mass and volume calculations. Use the gas law (22.4 L/mol at STP) consistently.

🎯 Exam Tip: Always start with the balanced chemical equation and use it as your roadmap for all stoichiometric calculations. Double-check that your mole ratios match the equation coefficients.

Solution 49.
Answer: a.
i. Atomic mass: S = 32 and O = 16
Molecular mass of SO₂ = 32+(2×16) = 64g
As 64 g of SO₂ = 22.4dm³

Then, 320 g of SO₂ = \( \frac{320 \times 22.4}{64} \) = 112 L

ii. Gay-Lussac's law Gay-Lussac's Law states "When gases react, they do so in volumes which bear a simple ratio to one another and to the volume of the gaseous product, if all the volumes are measured at the same temperature and pressure."

iii. C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Molar mass of propane = 44
44 g of propane requires 5 × 22.4 litres of oxygen at STP.
8.8 g of propane requires \( \frac{5 \times 22.4 \times 8.8}{44} \) = 22.4 litres

b.
i.

Empirical formula = CH₂Br
n(Empirical formula mass of CH₂Br) = Molecular mass (2 × VD)
n(12 + 2 + 80) = 94 × 2
n = 2
Molecular formula = Empirical formula × 2
= (CH₂Br) × 2
= C₂H₄Br₂

ii. 10²² atoms of sulphur
6.022 × 10²³ atoms of sulphur will have mass = 32 g
10²² atoms of sulphur will have mass = \( \frac{32 \times 10^{22}}{6.022 \times 10^{23}} \) = 0.533 g

iii. 0.1 mole of carbon dioxide
1 mole of carbon dioxide will have mass = 44 g
0.1 mole of carbon dioxide will have mass = 4.4 g

📝 Teacher's Note: For empirical formula problems, teach students to organize data in a table format. Emphasize that the simplest whole number ratio is key to finding the empirical formula.

🎯 Exam Tip: When converting atoms to mass, always use Avogadro's number as the conversion factor. Show all steps clearly to avoid calculation errors.

Solution 50.
Answer: a.
P + 5HNO₃(dil.) → H₃PO₄ + H₂O + 5NO₂

i. Number of moles of phosphorus taken = \( \frac{9.3}{31} \) = 0.3 mol

ii. 1 mole of phosphorus gives 98 gm of phosphoric acid.
So, 0.3 mole of phosphorus gives (0.3 × 98) gm of phosphoric acid
= 29.4 gm of phosphoric acid

iii. 1 mole of phosphorus gives 112 L of NO₂ gas at STP.
So, 0.3 mole of phosphorus gives (112 × 0.3) L of
NO₂ gas at STP.
= 33.6 L of NO₂ gas at STP

b.
i. According to the equation
N₂(g) + 3H₂(g) → 2NH₃(g)
3 volumes of hydrogen produce 2 volumes of ammonia
67.2 litres of hydrogen produce \( \frac{2 \times 67.2}{3} \) = 44.8 L
3 volumes of hydrogen combine with 1 volume of ammonia.
67.2 litres of hydrogen combine with \( \frac{1 \times 67.2}{3} \) = 22.4L Nitrogen left = 44.8 - 22.4 = 22.4 litres

ii. 5.6 dm³ of gas weighs 12 g
1 dm³ of gas weighs = (12/56) gm
22.4 dm³ of gas weighs = (12/56 × 22.2) gm = 48g
Therefore, the relative molecular mass of gas = 48 gm.

iii. Molar mass of Mg(NO₃)₂.6H₂O
= 24 × (14 × 2) + (16 × 12) + (1 × 12) = 256 g
Mass percent of magnesium = \( \frac{24 \times 100}{256} \) = 9.37%

📝 Teacher's Note: In Haber process calculations, remind students that the limiting reagent determines product formation. Use volume ratios from the balanced equation to find remaining gases.

🎯 Exam Tip: For gas density problems, remember that 22.4 L at STP contains one mole of any gas. Use this to find molar mass from given density data.

Solution 51.
Answer: a.
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
2V          13V
i. 2 vols. of butane requires O₂ = 13 vols
90 dm³ of butane will require O₂ = \( \frac{13}{2} \times 90 \) = 585 dm³

ii. Molecular mass = 2 × Vapour density
So, molecular mass of gas = 2 × 8 = 16 g
As we know, molecular mass or molar mass occupies 22.4 litres.
That is,
16 g of gas occupies volume = 22.4 litres
So, 24 g of gas will occupy volume
= \( \frac{22.4}{16} \times 24 = 33.6 \) litres

iii. According to Avogadro's law, equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules.
So, molecules of nitrogen gas present in the same vessel = X

b.
2KClO₃ \( \xrightarrow{MnO_2} \) 2KCl + 3O₂
2V               2V     3V
i. 3 vols. of oxygen require KClO₃ = 2 vols.
So, 1 vol. of oxygen will require KClO₃ = \( \frac{2}{3} \) vols.
So, 6.72 litres of oxygen will require KClO₃
So, 1 vol. of oxygen will require KClO₃ = \( \frac{2}{3} \) vols.
So, 6.72 litres of oxygen will require KClO₃
= \( \frac{2}{3} \times 6.72 = 4.48 \) litres
22.4 litres of KClO₃ has mass = 122.5 g
So, 4.48 litres of KClO₃ will have mass
= \( \frac{122.5}{22.4} \times 4.48 = 24.5 \) g

ii. 22.4 litres of oxygen = 1 mole
So, 6.72 litres of oxygen = \( \frac{6.72}{22.4} \) = 0.3 moles
No. of molecules present in 1 mole of O₂
= 6.023 × 10²³
So, no. of molecules present in 0.3 mole of O₂
= 6.023 × 10²³ × 0.3
= 1.806 × 10²³

iii. Volume occupied by 1 mole of CO₂ at STP = 22.4 litres
So, volume occupied by 0.01 mole of CO₂ at STP = 22.4 × 0.01 = 0.224 litres

📝 Teacher's Note: Avogadro's law is fundamental - equal volumes of gases at same T and P contain equal molecules. Use this principle to solve gas stoichiometry problems step by step.

🎯 Exam Tip: When dealing with gas reactions, always work with volume ratios first, then convert to mass or moles as needed. Keep track of your units throughout the calculation.

Solution 52.

Answer:
a.
i. \( 2C_2H_2 + 5O_2 \rightarrow 4CO_2 + 2H_2O \)
2 moles of \( C_2H_2 \) = 4 moles of \( CO_2 \)
x dm³ of \( C_2H_2 \) = 8.4 dm³ of \( CO_2 \)

\( x = \frac{2 \times 8.4}{4} \)
=4.2 dm³ of \( C_2H_2 \)

ii. Empirical formula= \( X_2Y \)
Atomic weight (X)= 10
Atomic weight (Y)= 5
Empirical formula weight = (2 × 10) + 5 = 25

\( n = \frac{\text{Molecular weight}}{\text{Empirical formula weight}} \)

\( = \frac{2 \times V.D}{\text{Empirical formula weight}} \)

\( = \frac{2 \times 25}{25} \)

= 2
So, molecular formula = \( X_2Y \times 2 \) = \( X_4Y_2 \)

b.
i. A cylinder contains 68 g of ammonia gas at STP.
Molecular weight of ammonia = 17 g/mole
68 g of ammonia gas at STP =?
1 mole = 22.4 dm³
∴ 4 mole = 22.4 × 4 = 89.6 dm³
ii. 4 moles of ammonia gas is present in the cylinder.
iii. 1 mole = \( 6.023 \times 10^{23} \) molecules
4 moles = \( 24.092 \times 10^{23} \) molecules

📝 Teacher's Note: When solving stoichiometry problems, always start by writing balanced equations and use the mole ratio method. Show students how to set up proportions clearly to avoid calculation errors.

🎯 Exam Tip: In gas volume calculations at STP, remember that 1 mole of any gas = 22.4 dm³. Always convert mass to moles first, then to volume.

Solution 42.

Answer:
The formula of aluminium nitride is AlN.

The molecular mass = 41

So, the percentage of N = \( 14 \times 100/41 = 34.146\% \)

📝 Teacher's Note: Emphasize the importance of writing correct chemical formulas first before calculating percentage composition. Students often make errors in formula writing which affects all subsequent calculations.

🎯 Exam Tip: Always double-check your chemical formula and atomic masses from the periodic table. Round percentage to appropriate significant figures as shown in the question.

Solution 48.

Answer:
(i) Element % atomic mass atomic ratio simple ratio
C 4.8 12 \( \frac{4.8}{12} = 0.4 \) 1
Br 95.2 80 \( \frac{95.2}{80} = 1.2 \) 3

So, empirical formula is \( CBr_3 \)
(ii) Empirical formula mass = 12 + 3 × 80 = 252 g
molecular formula mass = 2 × 252(V.D) = 504 g
n= 504/252 = 2
so, molecular formula = \( C_2Br_6 \)

📝 Teacher's Note: When finding empirical formulas, teach students to always divide by the smallest ratio to get whole numbers. If ratios are not whole, multiply all by a common factor.

🎯 Exam Tip: Remember that molecular mass = 2 × vapor density. Calculate empirical formula mass carefully and use n = molecular mass/empirical formula mass to find molecular formula.

Solution 49.

Answer:
\( 2C_8H_{18} + 25O_2 \rightarrow 16CO_2 + 18H_2O \)
2 V 25 V 16 V 18 V
(i) 2 moles of octane gives = 16 moles of \( CO_2 \)
so, 1 mole octane will give = 8 moles of \( CO_2 \)
(ii) 1 mole \( CO_2 \) occupies volume = 22.4 litre
so, 8 moles will occupy volume = 8 × 22.4 = 179.2 litre
(iii) 1 mole \( CO_2 \) has mass = 44 g
so, 16 moles will have mass = 44 × 16 = 704 g
(iv) Empirical formula is \( C_4H_9 \).

📝 Teacher's Note: For combustion reactions, remind students that hydrocarbons always produce CO₂ and H₂O. Use the balanced equation to establish mole ratios for all calculations.

🎯 Exam Tip: In combustion problems, always balance the equation first. Use mole ratios from the balanced equation to find amounts of products formed.

Solution 50.

Answer:
(a) (i) element % atomic mass at. ratio simple ratio
C 14.4 12 1.2 1
H 1.2 1 1.2 1
Cl 84.5 35.5 2.38 2
Empirical formula = \( CHCl_2 \)
(ii) Empirical formula mass = 12+1+71= 84 g
Since molecular mass = 168 so, n = 2
so, molecular formula = \( (CHCl_2)_2 = C_2H_2Cl_4 \)

(b) (i) \( C + 2H_2SO_4 \rightarrow CO_2 + 2H_2O + 2SO_2 \)
1 V 2 V 1 V 2 V
196 g of \( H_2SO_4 \) is required to oxidized = 12 g C
So, 49 g will be required to oxidise = 49 x 12/196 = 3 g
(ii) 196 g of \( H_2SO_4 \) occupies volume = 2 x 22.4 litres
So, 49 g \( H_2SO_4 \) will occupy = 2 x 22.4 x 49/196 = 11.2 litre
i.e. volume of \( SO_2 \) = 11.2 litre

📝 Teacher's Note: In empirical formula calculations, teach students to organize data in a clear table format. Always check if the molecular formula is a multiple of the empirical formula using the n factor.

🎯 Exam Tip: When solving redox reactions with acids, use stoichiometry carefully. Set up proportions using the balanced equation to find required masses and volumes.