Frank Brothers Solutions for ICSE Class 10 Chemistry Chapter 5 Mole Concept And Stoichiometry

Chapter 5. Mole Concept And Stoichiometry

Solution 1:
Answer:
1. Gay-Lussac's law: It states that 'when gases react, they do so in volumes which bear a simple ratio to one another, and also to the volume of the gaseous product, provided all the volumes are measured at the same temperature and pressure'.
2. Avogadro's law: It states that 'Under the same conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules'.
In simple words: Gay-Lussac's law tells us that gas volumes combine in simple whole number ratios during reactions, while Avogadro's law says equal volumes of any gas at the same conditions contain the same number of molecules.

📝 Teacher's Note: Use simple balloon examples to demonstrate volume relationships - if one balloon of hydrogen reacts with one balloon of chlorine, they produce two balloons of hydrogen chloride gas. This visual helps students grasp the concept.

🎯 Exam Tip: Always mention "at same temperature and pressure" when stating these laws - this condition is crucial for full marks.

Solution 2:
Answer:
(a) Atomicity of a gas: The number of atoms in a molecule of a gas is called its atomicity.
For example: Monoatomic means a gas molecule containing one atom. Similarly diatomic corresponds to two atoms and triatomic corresponds to three atoms in a molecule of a gas.
(b)

In simple words: Atomicity simply means how many atoms are stuck together to form one molecule of a gas - like hydrogen gas has 2 atoms per molecule, so it's diatomic.

📝 Teacher's Note: Use building blocks or LEGO to show students how atoms combine to form molecules. This hands-on approach makes atomicity concepts more memorable.

🎯 Exam Tip: Remember that 2H means two separate hydrogen atoms while \( H_2 \) means one hydrogen molecule containing two atoms bonded together.

Solution 3:
Answer: When stating the volume of a gas, the pressure and temperature should also be given because the volume of a gas is highly susceptible to slight change in pressure and temperature of the gas.
In simple words: Gas volume changes dramatically with temperature and pressure changes, so these conditions must always be mentioned for accuracy.

📝 Teacher's Note: Demonstrate with a balloon in hot water versus cold water to show how gas volume changes with temperature - students will visibly see the expansion and contraction.

🎯 Exam Tip: Always specify STP (Standard Temperature and Pressure) conditions when discussing gas volumes in numerical problems.

Solution 4:
Answer:
a. The relative atomic mass of Cl atom is 35.5 a.m.u. because chlorine consists of a mixture of two isotopes of masses 35 and 37 in the ratio of 3:1.
The average of the isotopic masses is 35 × 3 + 37 × 4 = 35.5
b. The value of Avogadro's number is \( 6.023 \times 10^{23} \).
c. The value of molar volume of a gas at STP is \( 22.4 \, dm^3 \) (litre) or \( 22400 \, cm^3 \) (ml).
Concept Insight: a. Isotopes are atoms of same having same atomic number but different mass number.
c. One mole of any gaseous molecules occupy \( 22.4dm^3 \) at standard temperature and pressure (STP). This volume is known as molar volume.
In simple words: Chlorine's atomic mass is 35.5 because it's a mixture of two types of chlorine atoms, and one mole of any gas always takes up 22.4 litres of space at standard conditions.

📝 Teacher's Note: Use the analogy of a mixed bag of marbles - if you have 3 marbles weighing 35g each and 1 marble weighing 37g, the average weight explains why chlorine has a fractional atomic mass.

🎯 Exam Tip: Remember the magic numbers: Avogadro's number \( 6.023 \times 10^{23} \) and molar volume 22.4 L at STP - these appear in many calculations.

Solution 5:
Answer:
(a) Vapour density: It is Density of a gas, expressed as the mass of a given volume of the gas divided by the mass of an equal volume of a reference gas (such as hydrogen or air) at the same temperature and pressure.
(b) Molar volume: One mole of any gaseous molecules occupy \( 22.4dm^3 \) at standard temperature and pressure (STP). This volume is known as molar volume.
"The molar volume of a gas can be defined as the volume occupied by one mole of a gas at standard temperature and pressure."
(c) Relative atomic mass: "The relative atomic mass or atomic weight 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".
Relative atomic mass = Mass of 1 atom of the element/1/12 of the mass of one \( C^{12} \) atom.
(d) Avogadro's number: Avogadro's number is defined as the number of atoms present in 12g of \( C^{12} \) isotope i.e. \( 6.023 \times 10^{23} \) atoms.
• It is number of elementary units i.e. atoms, ions or molecules present in one mole of a substance.
• It is denoted by \( N_A \).
(e) Relative molecular mass: "The relative atomic mass (or molecular weight) of an element or a 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.
In simple words: These are all ways to measure and compare masses and volumes of different substances using standard references - like comparing weights using a standard kilogram.

📝 Teacher's Note: Connect these concepts to everyday measurements - just like we compare heights to a meter stick, chemists compare atomic masses to carbon-12 as the standard reference.

🎯 Exam Tip: Learn the exact definitions word-for-word, especially for relative atomic mass and Avogadro's number - examiners expect precise terminology.

Solution 6:
Answer:
(a) The main applications of Avogadro's law are:
• Explanation of Gay-Lussac's Law
• Determination of atomicity of gases
• Determination of the molecular formula of a gaseous compound.
• Establishes relationship between the relative vapour density of a gas and its relative molecular mass.
• Establishes the relationship between gram molecular weight and volume of a gas at STP
(b) Explanation of Gay-Lussac's Law: Gay-Lussac had experimentally determined that one volume of hydrogen and one volume of chlorine react to produce two volumes of hydrogen chloride gas.
According to Avogadro's law, if:
1 volume of hydrogen contains n molecules of the gas then 1 volume of chlorine also contains n molecules of the gas. Therefore 2 volume of hydrogen chloride contain 2n molecules of the gas.
\( H_2 + Cl_2 \rightarrow 2HCl \)
1vol 1 vol 2 vol (by Gay-Lussac)
n n 2n (By Avogadro)
but hydrogen and chlorine are diatomic.
So, 2 atoms + 2 atoms → 2 molecules
1 atom + 1 atom → 1 molecule
i.e. 1 molecule of hydrogen chloride is formed when 1 atom of hydrogen combines with 1 atom of chlorine. Thus Avogadro's law explains Gay-Lussac's law of combining volumes.
In simple words: Avogadro's law helps us understand why gases combine in simple volume ratios and allows us to figure out how many atoms are in gas molecules.

📝 Teacher's Note: Draw the reaction as circles representing molecules to show students visually how one hydrogen molecule plus one chlorine molecule makes two hydrogen chloride molecules.

🎯 Exam Tip: When explaining Gay-Lussac's law using Avogadro's law, always show the volume ratios and molecular ratios side by side for clarity.

Solution 7:
Answer:
1. Gram atom: "The quantity of the element which weighs equal to its gram atomic mass is called one gram atom of that element".
For example: The gram atomic mass of hydrogen is 1g. So, 1g of hydrogen is 1 gram atom of hydrogen.
2. Gram mole: "A sample of substance with its mass equal to its gram molecular mass is called one gram molecule of this substance or one gram mole".
For example: Gram molecular mass of oxygen is 32 g. So One gram mole of oxygen is 32g.
In simple words: Gram atom means taking exactly the atomic mass worth of grams of an element, and gram mole means taking the molecular mass worth of grams of a compound.

📝 Teacher's Note: Use actual weighing scales and samples to show students what one gram atom or gram mole looks like in practice - this makes abstract concepts concrete.

🎯 Exam Tip: Always include the definition phrase and a numerical example when answering questions about gram atom or gram mole.

Solution 8:
Answer:
(a) The relative molecular mass of Potassium chlorate (\( KClO_3 \)) is:
[atomic mass of 1 K atom + Atomic mass of 1 Cl atom + atomic mass of 3 O atoms]
39 + 35.5 + 16 × 3 = 122.5
(b) The relative molecular mass of Sodium acetate (\( CH_3COONa \)) is:
[atomic mass of 2 C atom + Atomic mass of 3 H atom + atomic mass of 2 O atoms + atomic mass of 1 Na atom]
12×2 + 3 × 1 + 16 × 2 + 23 = 82
(c) The relative molecular mass of Chloroform (\( CH_3Cl \)) is:
[atomic mass of 1 C atom + Atomic mass of 3 H atom + atomic mass of 1 Cl]
12 + 3 ×1 + 35.5 = 50.5
(d) The relative molecular mass of Ammonium sulphate (\( (NH_4)_2SO_4 \)) is:
[atomic mass of 2 N atom + Atomic mass of 8 H atom + atomic mass of 1 S + atomic mass of 4 O atom]
14×2 + 8 × 1 + 32 + 16 × 4 = 132
In simple words: To find molecular mass, just add up the atomic masses of all atoms in the formula - like adding up the weights of all ingredients in a recipe.

📝 Teacher's Note: Teach students to first count each type of atom carefully, then multiply by atomic mass - use colored pens to highlight different elements in complex formulas.

🎯 Exam Tip: Always show your working step-by-step: count atoms, multiply by atomic masses, then add - this prevents calculation errors and earns method marks.

Solution 9:
Answer:
Empirical formula: "Empirical formula of a compound is the formula which gives the number of atoms of different elements present in one molecule of the compound, in the simplest numerical ratio".
Molecular formula: "Molecular formula of a compound denotes the actual number of atoms of different elements present in one molecule of the compound".
In simple words: Empirical formula shows the simplest ratio of atoms, while molecular formula shows the exact number of atoms in one molecule.

📝 Teacher's Note: Use the analogy of a recipe - empirical formula is like saying "1 part flour to 2 parts water" while molecular formula specifies "1 cup flour to 2 cups water" for exact quantities.

🎯 Exam Tip: Remember that molecular formula is always a whole number multiple of the empirical formula - this relationship often appears in numerical problems.

Solution 10:
Answer:
1. The empirical formula of \( C_6H_6 \) is: CH
2. The empirical formula of \( C_6H_{12}O_6 \) is: \( CH_2O \).
3. The empirical formula of \( C_2H_2 \) is: CH
4. The empirical formula of \( CH_3COOH \) is: \( CH_2O \).
In simple words: To find empirical formula, divide all subscripts by their highest common factor to get the simplest whole number ratio.

📝 Teacher's Note: Show students how to find the greatest common divisor of subscripts - practice with simple examples before moving to complex molecules.

🎯 Exam Tip: Always check if your empirical formula can be simplified further - look for common factors in all subscripts.

Solution 11:
Answer: Three pieces of information conveyed by the formula \( H_2O \) is that:
1. It shows that there are 2 hydrogen atoms and 1 oxygen atoms present in \( H_2O \).
2. The hydrogen and oxygen atoms are present in simplest whole number ratio of 2:1.
3. It represents one molecule of compound water.
In simple words: The formula \( H_2O \) tells us exactly what water is made of, in what ratio, and that it represents one complete water molecule.

📝 Teacher's Note: Use molecular models or drawings to show students how the formula translates to the actual 3D structure of the water molecule.

🎯 Exam Tip: When asked about information from a formula, always mention: composition, ratio, and that it represents one molecule - these are the three key points examiners look for.

Solution 12:
Answer: "A mole is defined as the amount (mass) of a substance containing elementary particles like atoms, molecules or ions equal to Avogadro's number i.e. \( 6.023 \times 10^{23} \).
Or a collection of \( 6.023 \times 10^{23} \) particles is called mole.
Number of elementary units present in one mole of a substance is \( 6.023 \times 10^{23} \).
In simple words: A mole is like a dozen - just as a dozen means 12 items, a mole means \( 6.023 \times 10^{23} \) particles of anything.

📝 Teacher's Note: Compare mole to familiar counting units like dozen, gross, or ream - this helps students understand that mole is simply a very large counting unit for tiny particles.

🎯 Exam Tip: Always include Avogadro's number \( 6.023 \times 10^{23} \) when defining mole - this specific number is crucial for full marks.

Solution 13:
Answer:
(a) Vapour density.
(b) One mole of gas.
(c) quantity of element which weighs equal to its gram atomic mass
(d) one
(e) \( 6.023 \times 10^{23} \)
In simple words: These are different ways to express quantities and properties of substances - like different units of measurement for different purposes.

📝 Teacher's Note: Create a reference chart showing the relationships between all these terms - students often confuse similar-sounding concepts like gram atom, gram mole, and molar mass.

🎯 Exam Tip: Learn these definitions as exact word matches - chemistry terminology is very precise and partial answers may not get full marks.

Solution 14:
Answer:
(a) Since one Oxygen molecule contains 2 atoms of oxygen, so it is a diatomic molecule.
O + O → \( O_2 \)
(b) The molecular mass of the given compound is determined experimentally by vapour density method also, in which the vapour density of the compound is determined. Vapour density is related to molecular mass as:
Molecular mass = 2 × vapour density.
(c) Since by definition of a mole it is defined as the amount (mass) of a substance containing elementary particles like atoms, molecules or ions equal to Avogadro's number i.e. \( 6.023 \times 10^{23} \) so one mole of any gas contains the same number of molecules.
In simple words: Oxygen is diatomic because two atoms stick together, molecular mass can be found using vapour density, and one mole always contains the same number of particles regardless of the substance.

📝 Teacher's Note: Emphasize that the key insight is "same number of particles" - whether it's hydrogen, oxygen, or any gas, one mole always contains Avogadro's number of particles.

🎯 Exam Tip: Remember the relationship: Molecular mass = 2 × vapour density - this formula is frequently tested in numerical problems.

Solution 15:
Answer:
1. \( Na_2SO_4.10H_2O \).
2. \( C_6H_{12}O_6 \).
In simple words: These are molecular formulas showing the exact composition of sodium sulfate hydrate and glucose respectively.

📝 Teacher's Note: Point out that the first compound shows hydrated salts (with water molecules attached) while the second shows a simple organic compound - different types of chemical formulas serve different purposes.

🎯 Exam Tip: When writing hydrated compound formulas, always include the dot and water molecules - \( Na_2SO_4.10H_2O \) shows this compound contains 10 water molecules per formula unit.

Solution 16:
Answer: Given reaction is: \( N_2 + O_2 \rightarrow 2NO \)
According to Gay-Lussac's law in the above reaction 1 volume of nitrogen combines with 1 volume of oxygen to produce 2 volumes of nitric oxide.
i.e. \( N_2 + O_2 \rightarrow 2NO \)
1 vol. 1 vol. 2 vol.
The volume of nitric oxide produced is = 1400 cm³.
Let the volumes of nitrogen and oxygen gases be = x
Then, \( N_2 + O_2 \rightarrow 2NO \)
x x 1400 cm³
So, x + x = 1400
2x = 1400
x = 1400/2 = 700 cm³
Hence the volumes of reacting gases i.e. nitrogen and oxygen is 700 cm³ each.
In simple words: When nitrogen and oxygen combine to form nitric oxide, they always react in equal volumes - like a 1:1 recipe that produces twice as much product.

📝 Teacher's Note: Use gas syringes or balloons to demonstrate gas volume relationships in chemical reactions. Students often confuse mass ratios with volume ratios - emphasize that for gases at same conditions, volume ratios equal mole ratios.

🎯 Exam Tip: Always write the balanced equation first, then use the molar volume concept - at STP, 1 mole of any gas occupies 22.4 L.

Solution 17:
Answer: Number of molecules in 12.8 g of sulphur dioxide gas.
Molecular mass of \( SO_2 \) = 64 a.m.u.
So, 64 g = 1 mole
12.8 g = 12.8/64 = 0.2 mole
Now 1 mole of \( SO_2 \) contains = \( 6 \times 10^{23} \) molecules
0.2 mole of \( SO_2 \) contains = \( 0.2 \times 6 \times 10^{23} \)
= \( 1.2 \times 10^{23} \) molecules.
In simple words: First find how many moles you have, then multiply by Avogadro's number to get the actual count of molecules.

📝 Teacher's Note: Students often forget the two-step process - moles first, then molecules. Practice with different compounds and masses to build familiarity with Avogadro's number.

🎯 Exam Tip: Always show the conversion from grams to moles clearly, then multiply by 6.022 × 10²³ - this earns method marks even if calculation is slightly wrong.

Solution 18:
Answer: Weight of \( 6 \times 10^{23} \) molecules of oxygen = 32g
Weight of 1 molecule of oxygen = \( 32/6 \times 10^{23} = 5.33 \times 10^{23} \)g
In simple words: If 6 × 10²³ molecules weigh 32g, then one single molecule weighs an incredibly tiny amount - about 5.33 × 10⁻²³ grams.

📝 Teacher's Note: This helps students understand how incredibly small individual molecules are. Compare this mass to everyday objects to give perspective on molecular scale.

🎯 Exam Tip: Remember that 6.022 × 10²³ molecules of any element equals its atomic mass in grams - this is the definition of Avogadro's number.

Solution 19:
Answer: Pb + \( (N)_2 + (O)_6 \)
207 + 2 × 14 + 6 × 16 = 331.
So, the molecular mass of \( Pb(NO_3)_2 \) = 331.
331 by weight of \( Pb(NO_3)_2 \) contain 96 parts by weight of oxygen.
100 parts will contain = 96 × 100/331 = 29%
So, the percentage composition of oxygen in lead nitrate is 29%.
In simple words: Out of every 331 grams of lead nitrate, 96 grams is oxygen - that's about 29% of the total weight.

📝 Teacher's Note: Emphasize the importance of correctly counting atoms in polyatomic ions. Students often forget to multiply by the subscript outside brackets.

🎯 Exam Tip: Always double-check your atomic mass calculations - count each type of atom carefully and multiply by the correct atomic masses.

Solution 20:
Answer: (a) Percentage of nitrogen in Ammonium nitrate \( [NH_4NO_3] \)
\( (N)_2 + (H)_4 + (O)_3 \)
14 × 2 + 1 × 4 + 3 × 16 = 80.
So, the molecular mass of \( NH_4NO_3 \) = 80.
80 by weight of \( NH_4NO_3 \) contain 28 parts by weight of nitrogen.
100 parts will contain = 28 × 100/80 = 35%
So, the percentage composition of nitrogen in Ammonium nitrate is 35%.
(b) Percentage of nitrogen in Ammonium phosphate \( [(NH_4)_3PO_4] \)
\( (N)_3 + (H)_{12} + P + (O)_4 \)
14 × 3 + 1 × 12 + 31 + 16 × 4 = 149.
So, the molecular mass of \( NH_4NO_3 \) = 149.
149 by weight of \( (NH_4)_3PO_4 \) contain 42 parts by weight of nitrogen.
100 parts will contain = 42 × 100/149 = 28.18%
So, the percentage of nitrogen in ammonium phosphate is 28.18%.
Since the percentage of nitrogen is more in Ammonium nitrate so it is a better fertilizer.
In simple words: Ammonium nitrate has 35% nitrogen while ammonium phosphate has only 28% nitrogen, so ammonium nitrate gives more nitrogen per gram to plants.

📝 Teacher's Note: This is a practical application students can relate to - fertilizers and farming. Explain why nitrogen percentage matters for plant growth.

🎯 Exam Tip: When comparing compounds, always calculate percentages and then clearly state which is better and why - this shows complete understanding.

Solution 21:
Answer:

Since the mole ratio for oxygen is fractional so we multiply the whole ratio by 3 to make it a whole number.
So, the empirical formula of the compound is \( Pb_3O_4 \).
In simple words: We convert percentages to mole ratios, then find the simplest whole number ratio - like a recipe showing how many atoms of each element combine.

📝 Teacher's Note: Students often struggle with fractional ratios. Teach them to multiply all ratios by the same number to get whole numbers - usually 2, 3, or 4 works.

🎯 Exam Tip: Always show the table format for empirical formula questions - it organizes your work and makes checking easier.

Solution 22:
Answer: Empirical formula of the compound is \( CH_2O \).
Empirical formula mass = Atomic mass of C + Atomic mass of H + Atomic mass of O
= 12 + 2 × 1 + 16 = 30.
Now as empirical formula is equal to the vapour density then;
Molecular mass = 2 × vapour density
= 2 × 30 = 60
n = Molecular mass / Empirical formula mass
= 60/30 = 2
Molecular formula = n × empirical formula
= 2 × \( (CH_2O) \)
= \( C_2H_4O_2 \)
The molecular formula of the compound is \( C_2H_4O_2 \).
In simple words: The empirical formula shows the simplest ratio, but the actual molecule is twice as big, so we double everything to get the molecular formula.

📝 Teacher's Note: Explain the relationship between vapour density and molecular mass clearly - many students confuse this. Use examples like water vapor to make it relatable.

🎯 Exam Tip: Remember: Molecular mass = 2 × Vapour density. This relationship is frequently tested and often forgotten under exam pressure.

Solution 23:
Answer: (i) From the equation:
Molecular weight of \( KNO_3 \) = (Atomic mass of K + Atomic mass of N + Atomic mass of O) = (39 + 14 + 16 × 3) = 101
Molecular mass of \( KNO_2 \) = (39 + 14 + 16 × 2) = 85
From the reaction:
2 moles of \( KNO_3 \) gives = 2 moles of \( KNO_2 \)
So, 202 g of \( KNO_3 \) gives = 170 g of \( KNO_2 \)
(ii) Given equation is: \( 2KNO_3 \rightarrow 2KNO_2 + O_2 \)
Molecular mass of \( KNO_3 \) is: (Atomic mass of K + Atomic mass of N + Atomic mass of O) = (39 + 14 + 16 × 3) = 101
Molecular mass of \( KNO_2 \) = (39 + 14 + 16 × 2) = 85
Now, decomposition of 101 g of \( KNO_3 \) yield = 16 g of \( O_2 \)
So, decomposition of 5.05 g of \( KNO_3 \) will yield = 16 × 5.05/101 = 0.8 g
Hence, when 5.05 g of potassium nitrate decomposes completely 0.8 g of oxygen is formed.
In simple words: When potassium nitrate breaks down, it loses some oxygen - we can calculate exactly how much oxygen gas is produced from any amount of the starting material.

📝 Teacher's Note: This demonstrates stoichiometry in decomposition reactions. Show students how to set up proportions using the balanced equation.

🎯 Exam Tip: Always balance the equation first, then use molar mass ratios to find the answer - this method works for any stoichiometry problem.

Solution 24:
Answer: (a) (i) Volume occupied by 48 g of oxygen.
As 32g of oxygen at STP occupies volume of = 22.4 L
48 g of oxygen at STP occupies volume of = 22.4 × 48/32 = 33.6 L
Hence, 48 g of sulphur dioxide will occupy a volume of 33.6 L
(ii) Volume occupied by 16 g of sulphur dioxide.
64 g of sulphur dioxide at STP occupies volume of = 22.4 L
16 g of sulphur dioxide at STP occupies volume of = 22.4 × 16/64 = 5.6 L
Hence, 16 g of sulphur dioxide will occupy a volume of 5.6 L
(b) 4 L of a gas at STP has mass = 5 g
22.4 L of a gas at STP will has molecular mass = 5 × 22.4/4 = 28
So, the molecular mass of a gas will be 28.
In simple words: At standard conditions, we can easily calculate gas volumes using the fact that 1 mole of any gas occupies 22.4 liters.

📝 Teacher's Note: Emphasize that 22.4 L/mol only applies at STP (Standard Temperature and Pressure). Use this as a gateway to introduce gas laws.

🎯 Exam Tip: Always mention STP conditions when using 22.4 L/mol - examiners look for this awareness of when formulas apply.

Solution 25:
Answer: Molecular formula of \( MgCO_3 \) is = 84
Molecular formula of \( H_2SO_4 \) = 98
Now if, 84 g of \( MgCO_3 \) requires = 98 g of \( H_2SO_4 \)
3 g of \( MgCO_3 \) will require = 98 × 3/84 = 3.5 g
So, 3.5 g of sulphuric acid will be required to dissolve 3 g of magnesium carbonate.
In simple words: We use the balanced chemical equation to find the exact amount of acid needed to completely dissolve a given amount of magnesium carbonate.

📝 Teacher's Note: This is a practical acid-base reaction students can observe in the lab. Demonstrate the fizzing of CO₂ gas when acid meets carbonate.

🎯 Exam Tip: Write the balanced equation first, then use the molar mass ratio - this systematic approach prevents calculation errors.

Solution 26:
Answer: Ferrous sulphate is \( FeSO_4 \cdot 7H_2O \)
Molecular mass of Ferrous sulphate is \( FeSO_4 \cdot 7H_2O \) is:
Atomic mass of Fe + Atomic mass of S + Atomic mass of H + Atomic mass of O
56 + 32 + 1 × 14 + 16 × 11 = 278
278 parts by weight of crystals contain 126 parts of water
100 parts will contain = 126 × 100/278 = 45.32%
So, the percentage of water in ferrous sulphate crystals is 45.32%
In simple words: Nearly half the weight of ferrous sulphate crystals is actually water molecules attached to the salt - this is called water of crystallization.

📝 Teacher's Note: Show students actual ferrous sulphate crystals and explain how heating removes the water, changing the crystal structure and color.

🎯 Exam Tip: Don't forget to include the water molecules (7H₂O) in your molecular mass calculation - this is a common mistake in hydrated salt problems.

Solution 27:
Answer:

So the Empirical formula of the compound will be \( CH_2Br \).
Now the Empirical formula mass will be = Atomic mass of C + Atomic mass of H + Atomic mass of Br
= 12 + 1 × 2 + 80 = 94
Now as Molecular mass = 2 × vapour density
= 2 × 94 = 188.
So n = Molecular mass / empirical formula mass
= 188/94 = 2
Molecular formula of the compound is = n × empirical formula
= 2 × \( (CH_2Br) \)
= \( C_2H_4Br_2 \)
In simple words: We found the simplest formula first, then used vapour density to find the actual molecular formula which is twice as large.

📝 Teacher's Note: Students often get confused between empirical and molecular formulas. Use simple examples like glucose (CH₂O vs C₆H₁₂O₆) to clarify the difference.

🎯 Exam Tip: Always show both empirical formula calculation and molecular formula calculation separately - they often carry separate marks in exams.

Solution 28:

Calculation of molar mass of M from first oxide:
Let us assume the atomic mass of M as x.
Atomic mass of Oxygen = 16

So, molar mass of M, x = 63.5

Calculation of formula of second oxide:

So formula of second oxide is M₂O.

Solution 29:

Conversion of sulphur dioxide to sulphur trioxide follows the following balanced equation:
4SO₂ + 2O₂→ 4SO₃
According to Gay-Lussac's law:
4 vol. 2 vol. 4 vol.
Volume of SO₂ = 4 vol. =300 mL
Volume of O₂ = 2 vol. =300 x 2 /4 = 150 mL
Volume of oxygen required = 21% = 150 mL
Volume of air required at STP = 100% = 100 x 150 /21 =714.28 mL
So, the volume of air at STP, required to convert 300 mL of sulphur dioxide to sulphur trioxide is 714.28 mL

Solution 30:

Given reaction is:
AgNO₃ + NaCl → AgCl + NaNO₃
Molecular mass of AgNO₃ is 170
Molecular mass of AgCl is 143.5
170 g of AgNO₃ produces =143.5 g of AgCl
17 g of AgNO₃ will produce = 143.5 x 17 /170 = 14.35 g
So, the weight of silver chloride precipitated is 14.35 g

Solution 31:

12 g of carbon contains = 6.023 x 10²³ number of carbon atoms
10⁻¹² g of carbon will contain = 6.023 x 10²³ x 10⁻¹² /12
= 0.5019 x 10¹¹
= 5.019 x 10¹⁰
So, the number of carbon atoms in the signature is 5.019 x 10¹⁰.

Solution 1996-1:

Calculation of number of molecules in each gas sample:
(i) 2L of carbon dioxide:
22.4 L of carbon dioxide has = 6.023 x 10²³ molecules
2 L of carbon dioxide will have = 6.023 x 10²³ x 2 /22.4 = 0.5377 x 10²³ molecules

(ii) 3L of chlorine
22.4 L of chlorine has = 6.023 x 10²³ molecules
3 L of chlorine will have = 6.023 x 10²³ x 3/22.4 = 0.8066 x10²³ molecules

(iii) 5 L of hydrogen
22.4 L of hydrogen has = 6.023 x 10²³ molecules
5 L of hydrogen will have = 6.023 x 10²³ x 5/22.4 = 1.34 x10²³ molecules

(iv) 4 L of nitrogen
22.4 L of nitrogen has = 6.023 x 10²³ molecules
4 L of nitrogen will have = 6.023 x 10²³ x 4/22.4 = 1.07 x10²³ molecules

(v) 1 L of sulphur dioxide
22.4 L of sulphur dioxide has = 6.023 x 10²³ molecules
1 L of sulphur dioxide will have = 6.023 x 10²³ x 1/22.4 = 0.27 x10²³ molecules

From the above calculation of number of molecules in different gases we can conclude that the:
(a) The greatest number of molecules are present in 5 L of hydrogen gas sample.
(b) The least number of molecules is in 1 L of sulphur dioxide gas sample.

Solution 1996-2:

(a) 64 g of SO₂ will be produced at STP from = 22.4L of H₂S
12.8 g of SO₂ will be produced at STP from = 22.4 x 12.8 /64
= 4.48 L
So, 4.48 L of hydrogen sulphide at STP will burn in oxygen to yield 12.8 g sulphur dioxide.

(b) From the equation we know that 2 moles of H₂S burn in presence of 3 moles of Oxygen so:
44.8 L of H₂S requires = 67.2 L of oxygen
4.48 L of H₂S will require = 67.2 /44.8 x 4.48 = 6.72 L
So, 6.72 L of oxygen would be required for complete combustion.

Solution 1996-3:

Molecular mass of Mg (NO₃)₂.6H₂O is = 256
Now, 256 parts by weight of crystal contains 192 parts by weight of oxygen.
So 100 parts by weight will contain = 192 x 100 /256 = 75%
Hence, total percentage of oxygen in magnesium nitrate crystal Mg(NO₃)₂.6H₂O is 75%.

Solution 1996-4:

Empirical formula of the compound is as:

So, the empirical formula of the compound is NH₂.

Solution 1997-1:

(a) No. it is not possible to change the temperature and pressure of a fixed mass of a gas without changing its volume because all the three variables are interrelated to each other by the gas equation as:
PV/T = K (constant) ------------------------------------------ 1 )
Hence if we change any one or two of the variables in the above equation then automatically third variable also has to change to make equation 1 equal to a constant.

(b) "The molar volume of a gas can be defined as the volume occupied by one mole of a gas at standard temperature and pressure". It has been noticed that one mole of any gaseous molecules occupy 22.4 L of volume at standard temperature and pressure.

Solution 1997-2:

Molecular mass of urea is =60
60 kg of urea has = 28 Kg of nitrogen
1000 Kg of urea will have = 28 x 1000 /60 = 466.66 or 467 Kg.

Solution 1997-3:

(a)

So the empirical formula of the compound is Na₂SiO₃.

(b) Given the empirical formula of the compound is C₂H₅.
Vapour density is = 29.
Empirical formula mass of the compound = 29
As, molecular mass = 2 x vapour density
= 2 x 29 = 58
So, molecular mass of the compound =58
Molecular formula = n x Empirical formula
Now, n = Molecular mass/ Empirical formula mass
= 58/29 =2
Molecular formula = 2 x (C₂H₅)
= C₄H₁₀
So, molecular formula of compound is = C₄H₁₀.

Solution 1997-4:

Molecular mass of ammonium dichromate = 252
Now, 252 g of ammonium dichromate evolves = 22.4 L of nitrogen at STP
63 g of ammonium dichromate will evolve = 22.4 x 63 /252 = 5.6 L
So, 63 g of ammonium dichromate will evolve 5.6 L of oxygen.

Solution 1998-1

Question. (a) Calculate the percentage of boron (B) in borax \( \text{Na}_2\text{B}_4\text{O}_7 \cdot 10\text{H}_2\text{O} \)

Answer: Molecular mass of borax \( \text{Na}_2\text{B}_4\text{O}_7 \cdot 10\text{H}_2\text{O} = 382 \) 382 parts by weight of borax contain 44 parts by weight of boron So 100 parts will contain = 44 × 100 / 382 = 11.5% Percentage of boron (B) in borax \( (\text{Na}_2\text{B}_4\text{O}_7 \cdot 10\text{H}_2\text{O}) = 11.5% \)

📝 Teacher's Note: Help students understand that they need to find the total atomic mass of all boron atoms (4 × 11 = 44) in the formula before calculating the percentage. Use the mole concept to reinforce this calculation.

🎯 Exam Tip: Always write the molecular formula clearly and show the calculation of molecular mass step by step to avoid arithmetic errors and get partial marks.

Question. (b) (i) Find the empirical formula of a compound with the following composition by mass: C = 26.7%, O = 71.1%, H = 2.2%

Answer:

So the empirical formula of the compound is \( \text{CO}_2\text{H} \)

📝 Teacher's Note: Show students how to set up the table systematically. Emphasize that they should divide all relative moles by the smallest value to get the simplest ratio.

🎯 Exam Tip: Present your work in a clear table format and always check that your final empirical formula makes chemical sense.

Question. (ii) The relative molecular mass of A is 90. Find the molecular formula if the empirical formula mass is 45.

Answer: Empirical Formula mass of compound is = 45 Then, n = Molecular mass/Empirical Formula Mass = 90/45 = 2 Molecular formula of compound = n × Empirical formula = 2 × \( (\text{CO}_2\text{H}) \) = \( \text{C}_2\text{O}_4\text{H}_2 \)

📝 Teacher's Note: Connect this to the concept that molecular formula is always a whole number multiple of the empirical formula. Use simple examples like \( \text{H}_2\text{O}_2 \) and \( \text{HO} \) to illustrate.

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

Solution 1998-2

Question. (i) Given equation is: \( 2\text{H}_2\text{O} \text{(l)} \rightarrow 2\text{H}_2 \text{(g)} + \text{O}_2 \text{(g)} \). Calculate the volume of oxygen produced.

Answer: According to Gay-Lussac's law: 2 volume of water produces 2 volume of hydrogen and 1 volume of oxygen i.e 2 volume of water produces = 2 volume of hydrogen = 2500 \( \text{cm}^3 \) 2 volume of water will produce = 1 volume of Oxygen = 2500/2 = 1250 \( \text{cm}^3 \) i.e. = 1250 \( \text{cm}^3 \)

📝 Teacher's Note: Emphasize that Gay-Lussac's law applies to gases at the same temperature and pressure. Use the balanced equation to find volume ratios directly.

🎯 Exam Tip: Always check that your balanced equation shows the correct mole ratios before applying Gay-Lussac's law for volume calculations.

Question. (ii) Given equation is: \( 2\text{NH}_3 \text{(g)} + 2\frac{1}{2}\text{O}_2 \text{(g)} \rightarrow 2\text{NO} \text{(g)} + 3\text{H}_2\text{O} \text{(l)} \). Find the mass of water produced from 1.5g of NO.

Answer: Molecular mass of NO = 30 Molecular mass of \( \text{H}_2\text{O} \) = 18 From the equation: 2 moles of NO = 3 moles of \( \text{H}_2\text{O} \) 60 g of NO = 54 g of \( \text{H}_2\text{O} \) 1.5g of NO = 54 × 1.5 / 60 = 1.35 g of \( \text{H}_2\text{O} \)

📝 Teacher's Note: Show students how to convert the fractional coefficient \( 2\frac{1}{2} \) to 2.5 or multiply the entire equation by 2 to avoid fractions.

🎯 Exam Tip: Set up the proportion correctly using the mole ratio from the balanced equation, then convert to mass using molecular masses.

Solution 1999-1

Question. (a) Given equation is: P + 5HNO₃ → H₃PO₄ + H₂O + 5NO₂. Calculate: (i) mass of phosphoric acid from 6.2g of phosphorous, (ii) mass of nitric acid consumed, (iii) volume of steam at 760mm Hg and 273°C

Answer: (i) Molecular mass of phosphorous = 31 Molecular mass of phosphoric acid = 98 31 g of phosphorous produces = 98 g of phosphoric acid 6.2 g of phosphorous will produce = 98 × 6.2 / 31 = 19.6 g Hence, 19.6 g of phosphoric acid can be prepared from 6.2 g of phosphorous. (ii) Molecular mass of nitric acid = 63 31 g of phosphorous will consume = 63 g of nitric acid. (iii) Moles of steam formed from 31g phosphorus = 1 mol moles of steam from 6.2g phosphorus = 1mol × 6.2g / 31g = 0.2 mol. volume of steam produced at S.T.P = (0.2 mol) × (22.4 L/mol) = 4.48 litre. Since the pressure (760 mm) remains constant, but the temperature (273 + 273) = 546 is doubled, the volume of the steam also gets doubled ∴ volume of steam produced at 760 mm Hg and 273°C = 4.48 × 2 = 8.96 litres.

📝 Teacher's Note: Break down stoichiometry problems into steps: balance equation → find mole ratios → convert between mass and moles → apply gas laws if needed.

🎯 Exam Tip: For gas volume calculations, first find the volume at STP, then apply gas laws to get volume at given conditions.

Question. (b) 4NH₃ + 5O₂ → 4NO + 6H₂O. Calculate volume of nitric oxide from 27 litres of reactants.

Answer: 4NH₃ + 5O₂ → 4NO + 6H₂O 4 vol. + 5 vol. = 4 vol. Volume of reactants = 4 vol. of ammonia + 5 vol. of oxygen = 9 vol. 9 vol. of reactants produces 4 vol. of Nitric oxide Therefore, 27 vol. of reactants will produce 4 × 27 lit / 9. = 12 litres of Nitric oxide

📝 Teacher's Note: Emphasize that in gas-phase reactions, volume ratios equal mole ratios when temperature and pressure are constant.

🎯 Exam Tip: Add up all reactant volumes first, then use proportion to find the product volume based on the balanced equation.

Solution 1999-2

Question. Calculate the mass of calcium nitrate required to replace nitrogen in a 10 hectare field containing 20 kg nitrogen per hectare.

Answer: Molecular weight of \( \text{Ca}(\text{NO}_3)_2 \) = 164 164 parts by weight of calcium nitrate contains 28 parts by weight of nitrogen. 28 Kg of nitrogen will be replaced by = 164 Kg of \( \text{Ca}(\text{NO}_3)_2 \) 20 Kg of nitrogen will be replaced by = 164 × 20 / 28 = 117.14 kg For, 1 hectare of field 20 Kg of nitrogen will be replaced by = 117.14 Kg of \( \text{Ca}(\text{NO}_3)_2 \) For, 10 hectare of field 20 Kg of nitrogen will be replaced by = 117.14 × 10 = 1171.4 Kg Hence, 1171.4 Kg of the fertilizer calcium nitrate, \( \text{Ca}(\text{NO}_3)_2 \) would be required to replace nitrogen in 10 hectare field.

📝 Teacher's Note: Help students identify that calcium nitrate contains 2 nitrogen atoms, so the nitrogen mass is 2 × 14 = 28. This is a common mistake area.

🎯 Exam Tip: First calculate the fertilizer needed for 1 hectare, then multiply by the total area to avoid confusion.

Solution 1999-3

Question. (a) Explain using Avogadro's law why equal volumes of oxygen and sulphur dioxide contain the same number of molecules at the same temperature and pressure.

Answer: As we know from Avogadro's law that under same conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules so if a vessel contains N molecules of oxygen at a certain temperature and pressure and since the vessel remains same so same volume of sulphur dioxide will be present in the vessel and hence N molecules of sulphur dioxide can be accommodated in the vessel at the same temperature and pressure.

📝 Teacher's Note: Use concrete examples with actual numbers (like 6.023 × 10²³ molecules in 22.4L at STP) to make Avogadro's law more tangible for students.

🎯 Exam Tip: State Avogadro's law clearly first, then apply it to the specific gases mentioned in the question.

Question. (b) (i) Compare the number of molecules in 2g of oxygen gas and 2g of hydrogen gas.

Answer: Flask having oxygen gas: 16 g of oxygen gas has = \( 6.023 \times 10^{23} \) molecules 2g of oxygen gas will have = \( 6.023 \times 10^{23}/16 \times 2 = 0.75 \times 10^{23} \) Flask having hydrogen: 1 g of hydrogen has = \( 6.023 \times 10^{23} \) molecules 2g of hydrogen gas will have = \( 6.023 \times 10^{23} \times 2 / 1 = 12.05 \times 10^{23} \) So, hydrogen gas has greater number of molecules.

📝 Teacher's Note: Emphasize that lighter molecules have more particles for the same mass. Use this to connect to the concept of molar mass and Avogadro's number.

🎯 Exam Tip: Always start with the relationship: 1 mole of any gas = 6.023 × 10²³ molecules, then convert mass to moles.

Question. (ii) If equal amounts (by mass) of oxygen and hydrogen gases are present, compare the number of molecules.

Answer: Amount of hydrogen and oxygen gases is same = 2 g So, for oxygen 32 g of gas has = N molecules Then, 2g of gas has = N/32 × 2 = 16 No of molecules of oxygen = N/16.

📝 Teacher's Note: This builds on part (i) - when masses are equal, the gas with lower molecular mass will have more molecules.

🎯 Exam Tip: Set up the calculation clearly showing that for equal masses, number of molecules is inversely proportional to molecular mass.

Solution 2000-2

Question. Find the empirical formula of a compound with N = 42%, O = 48%, H = 9%

Answer:

So the empirical formula of the compound is \( \text{NH}_3\text{OH} \).

📝 Teacher's Note: Point out that this is actually hydroxylamine (NH₂OH), but the empirical formula calculation gives the simplest ratio regardless of structure.

🎯 Exam Tip: Double-check that your percentages add up to 100% before starting the calculation to catch any errors early.

Solution 2000-1

Question. (a) At STP, calculate the mass of sulphur produced from H₂S and Cl₂

Answer: At STP, 22400 \( \text{cm}^3 \) of each H₂S and Cl₂ will give 32 g of sulphur i.e. 44800 \( \text{cm}^3 \) of H₂S + Cl₂ gives = 32 g of S (112 + 120) = 232 \( \text{cm}^3 \) of H₂S + Cl₂ will give = 32 × 232 / 44800 = 0.16 g of S

📝 Teacher's Note: Show the balanced equation H₂S + Cl₂ → 2HCl + S first, so students understand the 1:1 mole ratio.

🎯 Exam Tip: Always state the balanced chemical equation before doing stoichiometric calculations to show your understanding.

Question. (b) Calculate the mass of sodium carbonate formed from washing soda on heating

Answer: \( \text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O} \xrightarrow{\Delta} \text{Na}_2\text{CO}_3 + 10 \text{H}_2\text{O} \) molecular weight of washing soda is 286.14 g molecular weight of sodium carbonate is 106 g 286.14 g of \( \text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O} \) forms 106 g of sodium carbonate on heating 57.2 g of \( \text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O} \) forms = 106 × 57.2 / 286.14 = 21.2 g

📝 Teacher's Note: Explain that heating removes only the water of crystallization, not the carbonate itself. This is a dehydration reaction, not decomposition.

🎯 Exam Tip: Be careful to write the correct formula for washing soda with 10 water molecules, not the anhydrous form.

Question. (c) Calculate the mass of lead sulphate and sodium sulphate formed

Answer: \( \text{Na}_2\text{SO}_4 + \text{Pb}(\text{NO}_3)_2 \rightarrow \text{PbSO}_4 + 2\text{NaNO}_3 \) molecular weight of \( \text{Na}_2\text{SO}_4 \) is 142 g molecular weight of \( \text{PbSO}_4 \) is 303 g 303 g of \( \text{PbSO}_4 \) is formed by 142 g of \( \text{Na}_2\text{SO}_4 \) 15.1 g of \( \text{PbSO}_4 \) is formed by = 142 × 15.1 / 303 = 7.1 g of \( \text{Na}_2\text{SO}_4 \)

📝 Teacher's Note: This appears to be asking for reactant mass from product mass - a reverse stoichiometry problem. Make sure students can work in both directions.

🎯 Exam Tip: When working backwards from product to reactant, set up the proportion carefully and double-check your calculation.

Solution 2000-1: Gay-Lussac proposed this law.

Solution 2001-2: Molecular mass of ethane = 30. According to Gay-Lussac's law: 2 vol. of C₂H₆ requires = 7 vol. of oxygen. Vol. of C₂H₆ = 2 vol. = 100 L. Vol. of oxygen required = 7 vol. = 350 L

Solution 2001-3

Question. (a) Calculate the number of molecules in different gases at the same volume

Answer: (a) If 20 L of nitrogen has = X number of molecules Then, 10 L of chlorine will have = X × 10 / 20 = X/2. (b) If 20 L of nitrogen has = X number of molecules Then, 20 L of ammonia will have = X × 20 / 20 = X. (c) If 20 L of nitrogen has = X number of molecules Then, 5 L of sulphur dioxide will have = X × 5 / 20 = X/4.

📝 Teacher's Note: This directly applies Avogadro's law - equal volumes contain equal numbers of molecules. Use this to reinforce proportional thinking.

🎯 Exam Tip: Set up simple proportions based on volume ratios since the number of molecules is directly proportional to volume at constant temperature and pressure.

Solution 2001-4: The term is vapour density.

Solution 2001-5

Question. Calculate the volume of dinitrogen oxide required to produce steam

Answer: According to Gay-Lussac's law: In the equation \( 4 \text{N}_2\text{O} + \text{CH}_4 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} + 4\text{N}_2 \) Vol. of \( \text{H}_2\text{O} \) produced is = 2 vol. = 150 \( \text{cm}^3 \) Vol. of \( \text{N}_2\text{O} \) required is = 4 vol. = 150 × 4 / 2 = 300 \( \text{cm}^3 \) 300 \( \text{cm}^3 \) of dinitrogen oxide \( (\text{N}_2\text{O}) \) is required to give 150 \( \text{cm}^3 \) of steam.

📝 Teacher's Note: Point out that water is produced as steam (gas phase) in this high-temperature reaction, so volume ratios apply directly.

🎯 Exam Tip: Check the physical states in the equation - Gay-Lussac's law only applies when all substances are in the gas phase.

Solution 2001-6

Question. Calculate the percentage of phosphorous in superphosphate fertilizer

Answer: Molecular mass of fertilizer superphosphate, \( \text{Ca}(\text{H}_2\text{PO}_4)_2 \) = 234 234 parts by weight of fertilizer contains 62 parts by weight of phosphorous So, 100 parts will contain = 62 × 100 / 234 = 26.5%

📝 Teacher's Note: Help students identify that there are 2 phosphorous atoms in the formula, so the total P mass is 2 × 31 = 62.

🎯 Exam Tip: Always count the number of atoms of the element you're looking for in the compound formula before calculating mass.

Solution 2001-7

Question. Find the molecular formula of a metal chloride with density relative to hydrogen = 162.5 and percentage of chlorine = 65.5%

Answer: Given, density of chloride relative to hydrogen = 162.5 Percentage of chlorine = 65.5% Percentage of Metal M = 100 - 65.5 = 34.5%

Empirical formula = \( \text{MCl}_3 \) Empirical formula mass = 56 + 3 × 35.5 = 162.5 Molecular mass = 2 × Vapour density = 2 × 162.5 = 325 n = Molecular mass / Empirical formula mass = 325/162.5 = 2 Molecular formula = n × empirical formula = 2 × \( \text{MCl}_3 \) = \( \text{M}_2\text{Cl}_6 \)

📝 Teacher's Note: Explain that vapour density is half the molecular mass. Students often confuse this relationship, so practice with several examples.

🎯 Exam Tip: Remember the formula: Molecular mass = 2 × Vapour density. This is a key relationship in vapour density problems.

Solution 2002-1:

Answer: (a) X molecules of \( N_2 \) occupies V litres. 3X molecules of CO occupies 3V litres. (b) X molecules of \( O_2 \) = 8/32 = 1/4 mole of \( O_2 \) So, X molecules of \( CO_2 \) = 1/4 molecule of \( CO_2 \) So, Mass of \( CO_2 \) present in the sample = ¼ x gram molecular mass of \( CO_2 \) = 1/4 x 44 = 11g (c) Avogadro's law.

📝 Teacher's Note: Use Avogadro's law to connect equal volumes containing equal numbers of molecules at same conditions. Help students visualize this with gas balloons of equal size.

🎯 Exam Tip: Always state Avogadro's law when dealing with gas volume and molecule relationships - it's the key concept examiners look for.

Solution 2002-2:

Answer: Molecular formula of ammonium chloroplatinate \( (NH_4)_2PtCl_6 \): 2 x (atomic mass of N + 8 x atomic mass of H) + atomic mass of platinum + 6 x atomic mass of chlorine 2 x (14 + 8) + 195 + 6 x 35.3= 444 444 parts of ammonium chloroplatinate contains 195 parts by weight of platinum So, 100 parts will contain = 195 x 100/444 = 43.9% = 44%

📝 Teacher's Note: Emphasize the step-by-step approach: find molecular mass first, then identify the mass of the desired element, finally calculate percentage. Use real examples like finding iron content in tablets.

🎯 Exam Tip: Always show the molecular mass calculation clearly and double-check your arithmetic - percentage composition questions have multiple calculation steps where errors can accumulate.

Solution 2002-3:

Answer: Empirical formula = \( Na_3PO_4 \)

📝 Teacher's Note: Teach students to always divide by the smallest mole ratio to get the simplest whole number ratio. Practice with compounds having different complexity levels.

🎯 Exam Tip: Present your work in a clear table format - it makes checking easier and shows organized thinking to the examiner.

Solution 2003-1:

Answer: (a) \( CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \) 1 vol. 2 vol. 1 vol. 2 vol. One volume of methane requires oxygen = 2 vol. So, Vol. of oxygen used = 2 x 22.4 = 44.8 \( dm^3 \) \( 2H_2 + O_2 \rightarrow 2H_2O \) 2 vol. 1vol. 2 vol. 2 volume of hydrogen needs one volume of oxygen So, Volume of oxygen used =22.4 / 2 = 11.2 \( dm^3 \) Total volume of oxygen used = 44.8 + 11.2 = 56.0 \( dm^3 \) (b) Calculation of number of molecules in each gas sample: (i) 8 g of hydrogen: 1 g of hydrogen = \( 6.023 \times 10^{23} \) molecules 8 g of hydrogen will have = \( 6.023 \times 10^{23} \times 8 = 48.184 \times 10^{23} \) molecules (ii) 8 g of oxygen: 32 g of oxygen has = \( 6.023 \times 10^{23} \) molecules 8 g of oxygen will have = \( 6.023 \times 10^{23}/32 \times 8 = 1.50 \times 10^{23} \) molecules (iii) 8 g of carbon dioxide: 44 g of carbon dioxide has = \( 6.023 \times 10^{23} \) molecules 8 g of carbon dioxide will have = \( 6.023 \times 10^{23}/44 \times 8 = 1.0 \times 10^{23} \) molecules (iv) 8 g of sulphur dioxide: 64 of sulphur dioxide has = \( 6.023 \times 10^{23} \) molecules 8 g of sulphur dioxide will have = \( 6.023 \times 10^{23}/64 \times 8 = 0.75 \times 10^{23} \) molecules (v) 8 g of chlorine: 35.5 g of chlorine has = \( 6.023 \times 10^{23} \) molecules 8 g of chlorine will have = \( 6.023 \times 10^{23}/71 \times 8 = 0.68 \times 10^{23} \) molecules Hence, chlorine gas will have least number of molecules. Hydrogen gas will have the most number of molecules.

📝 Teacher's Note: Help students understand that equal masses of different gases contain different numbers of molecules because molecular masses are different. Use the analogy of different sized marbles in equal-weight bags.

🎯 Exam Tip: Always calculate systematically and compare final values clearly. State your conclusion explicitly - don't make the examiner hunt for your answer.

Solution 2004-1:

Answer: (a) (i) Moles = Weight of substance in grams/ molecular weight So, Moles of \( SO_2 \) = 3.2 /64 = 0.05 (ii)Number of molecules = Moles x \( 6.023 \times 10^{23} \) So, number of molecules of \( SO_2 \) = \( 0.05 \times 6.023 \times 10^{23} \) = \( 0.30115 \times 10^{23} = 0.302 \times 10^{23} \) (iii) 64 g of \( SO_2 \) occupy a volume = 22.4 L So, 3.2 g of \( SO_2 \) will occupy a volume = \( 22.4 \times 3.2 /64 = 1.12 L \) (b) Molecular weight of \( KMnO_4 \) = 39 + 55 + 16 x 4 = 158 Molecular weight of \( K_2SO_4 \) = 2 x 39 + 32 + 16 x 4 = 174 Molecular weight of \( FeSO_4 \) = 56 + 32 + 64 = 152 2 x 158 g of \( KMnO_4 \) yields = 174 g of \( K_2SO_4 \) So, 15.8 g of \( KMnO_4 \) will yield = \( 174 \times 15.8 / 2 \times 158 = 8.7 \) g of \( K_2SO_4 \) 174 g of \( K_2SO_4 \) yields 152 g of \( FeSO_4 \) So, 8.7 g of \( K_2SO_4 \) will yield = \( 152 \times 8.7/174 = 7.6 \) g of \( FeSO_4 \) Hence, 7.6 g of iron (II) sulphate is used in the above reaction.

📝 Teacher's Note: Break down stoichiometry problems into clear steps: balance equation, find molar masses, use ratios. Practice with simpler examples before complex multi-step problems.

🎯 Exam Tip: Show all molecular mass calculations explicitly and use unitary method clearly - this demonstrates systematic thinking and reduces errors.

Solution 2004-2:

Answer: (a) Mass of one litre of oxygen gas liberated at room temperature = 1.32 g Mass of one litre of hydrogen under the same conditions of temperature and pressure = 0.0825 g Relative Molecular mass of oxygen = Weight of n molecule of \( O_2 \)/Weight of 1/2 molecule of hydrogen = 1.32X2/0.0825 = 32 a.m.u. (b) \( 2 KMnO_4 \rightarrow K_2MnO_4 + MnO_2 + O_2 \) Molecular mass of \( KMnO_4 \) = 158 Molar volume of \( O_2 \) at room temperature = 24 L 2 x158 g of \( KMnO_4 \) at room temperature yields = 24 L of \( O_2 \) 15.8 g of \( KMnO_4 \) will yield = \( 24 \times 15.8 / 2 \times 158 = 1.2 L \) of \( O_2 \).

📝 Teacher's Note: Emphasize that relative molecular mass is a comparison - always explain what it's relative to. Use everyday examples like comparing weights of different fruits.

🎯 Exam Tip: Remember that molar volume changes with temperature - use 22.4 L at STP and 24 L at room temperature as given in the problem.

Solution 2005-1:

Answer: (a) \( 2NaHCO_3 \rightarrow Na_2CO_3 + H_2O + CO_2 \) Molecular mass of \( NaHCO_3 \) = 84 Molecular mass of \( Na_2CO_3 \) =106 From the above reaction: 1 \( Na_2CO_3 \) is obtained from = 2 \( NaHCO_3 \) 106 g of \( Na_2CO_3 \) is obtained from = 168 g of \( NaHCO_3 \) So, 21.2 g of \( Na_2CO_3 \) will be obtained from = \( 168 \times 21.2 /106 = 33.6 \) g of \( NaHCO_3 \) (b) \( NaCl + NH_3 + CO_2 + H_2O \rightarrow NaHCO_3 + NH_4Cl \) From the equation, 1 mole of \( CO_2 \) i.e. 22.4 L of \( CO_2 \) is used to produce = 1 mole of \( NaHCO_3 \) Now as, 84 g of \( NaHCO_3 \) requires = 22.4 L of \( CO_2 \) 33.6 g of \( NaHCO_3 \) will require = \( 22.4 \times 33.6 /84 = 8.96 L \) of \( CO_2 \).

📝 Teacher's Note: Connect the two parts of the problem - show how the product from part (a) becomes the starting point for calculations in part (b). This demonstrates real industrial processes.

🎯 Exam Tip: Always check if your answer makes logical sense - 33.6g should need less CO₂ than 84g, which it does (8.96L < 22.4L).

Solution 2006-1:

Answer: (a) (i) Gas D contains the maximum number of molecules. (ii) If the temperature and the pressure of gas A are kept constant, then the volume of the gas will get doubled. (iii) Gay -Lussac's law of combining volumes. (iv) Gases A D 1 : 4 5.6 \( dm^3 \) 4 x 5.6 \( dm^3 \) at STP 22.4 \( dm^3 \) (molar volume) \( 6 \times 10^{23} \) molecules. (v)\( 6 \times 10^{23} \) molecules is Avogadro's number of molecules contained in one gram mole of the substance if gas D is \( N_2O \) then, 1 gram mole of \( N_2O \) = 2 x 14 + 16 = 44 g (b) Molecular mass of aluminium nitride \( (AlN_3) \) = 27 + 14 x 3 = 69 Now, 69 parts by weight of aluminium nitride contains = 42 parts by weight of nitrogen So, 100 parts will contain = \( 42 \times 100 /69 = 60.86% \) Hence, the percentage of nitrogen in aluminium nitride is 60.86%.

📝 Teacher's Note: Use gas law relationships to connect different properties. Draw diagrams showing how doubling molecules while keeping temperature and pressure constant affects volume.

🎯 Exam Tip: When dealing with gas problems, identify which gas law applies and state it clearly - Gay-Lussac's, Avogadro's, or combined gas law.

Solution 2006-2:

Answer: (a) Determination of empirical formula:

The empirical formula of compound = \( K_2BeF_4 \) (b) \( 3CuO + 2NH_3 \rightarrow 3Cu + 3H_2O + N_2 \) Molecular mass of CuO = 80 Volume occupied by 1 mole of \( NH_3 \) = 22.4 L From the equation: 3 moles of CuO is reduced by 2 moles of ammonia For 240 g of CuO, volume of \( NH_3 \) consumed = 44.8 L For 120 g of CuO, volume of \( NH_3 \) consumed = \( 44.8 \times 120 /240 = 22.4 L \)

📝 Teacher's Note: Emphasize the systematic approach for empirical formulas and stoichiometry. Students often rush and make calculation errors - encourage checking work step by step.

🎯 Exam Tip: Always verify your empirical formula makes chemical sense - check if the compound is reasonable and if your ratios are truly in lowest terms.

Solution 2006-3:

Answer: (a) Molecular mass of ethylene \( (CH_2-CH_2) \) = 28 g Number of moles = Given weight / Molecular weight = 1.4 / 28 = 0.05 moles Now, number of molecules in 1 mole = \( 6.023 \times 10^{23} \) So, number of molecules in 0.05 moles =\( 6.023 \times 10^{23} \times 0.05 = 0.3 \times 10^{23} \) = \( 3 \times 10^{22} \) molecules. Volume occupied by 1 mole of ethylene = 22.4 L So, volume occupied by 0.05 moles of ethylene = \( 22.4 \times 0.05 \) = 1.12 L (b) Vapour density = Molecular weight / 2 = 28 / 2 = 14.

📝 Teacher's Note: Connect molecular calculations to practical applications. Explain how vapour density relates to how gases behave in real life situations like hot air balloons.

🎯 Exam Tip: Remember the relationship: Vapour Density = Molecular Weight ÷ 2. This is a standard formula that appears frequently in exams.

Solution 2006-4

Question. (a) Molecular weight of sodium aluminium fluoride (\( Na_3AlF_6 \)) = 210
Answer: Now, 210 parts by weight of \( Na_3AlF_6 \) contains = 69 parts by weight of sodium
So, 100 parts will contain = 69 × 100 / 210 = 32.8 or 33%
In simple words: We find what percentage of the total compound weight is just sodium by using simple proportion.

📝 Teacher's Note: Show students how to identify the atomic mass of sodium (23) and count how many sodium atoms are in the formula (3), giving 3 × 23 = 69. This makes the calculation clearer.

🎯 Exam Tip: Always write the molecular weight breakdown first, then use proportion to find the percentage. Show your calculation steps clearly.

Question. (b) \( 2CO + O_2 → 2CO_2 \)
Answer: From the reaction:
2 volumes of CO consumes = 1 volume of \( O_2 \)
So, 560 mL of CO consumes = ½ × 560 = 280 mL
Now, 2 volume of CO gives = 2 volume of \( CO_2 \)
So, 560 mL of CO will give = 2 × 560 / 2 = 560 mL
In simple words: The reaction shows that the same volume of CO produces the same volume of carbon dioxide gas.

📝 Teacher's Note: Use Gay-Lussac's law here - equal volumes of gases at same conditions react in simple ratios. Draw the reaction with volume ratios above the equation.

🎯 Exam Tip: For gas volume calculations, always write the volume ratio from the balanced equation first, then apply proportion.

Solution 2007-1

Question. (i) \( NH_4NO_3 → N_2O + 2H_2O \)
Answer: From the equation:
1 mole of \( NH_4NO_3 \) yields 2 mole of \( H_2O \)
So, 44.8 L of steam = 22.4 L of \( N_2O \) at STP
8.96 L of steam = 22.4 × 8.96 / 44.8 = 4.48 L of \( N_2O \) at STP.
In simple words: We use the mole ratio from the equation to find how much nitrous oxide gas is produced from the given amount of steam.

📝 Teacher's Note: Emphasize that 22.4 L is the molar volume of any gas at STP. Students should memorize this value and understand its application.

🎯 Exam Tip: Always convert volumes to moles using 22.4 L/mol at STP, then use mole ratios from the balanced equation.

Question. (ii) Molecular mass of \( NH_4NO_3 \) = 80
Answer: 44.8 L of steam is liberated by = 80 g of \( NH_4NO_3 \)
8.96 L of steam will be liberated by = 80 × 8.96 / 44.8 = 16 L
In simple words: We find how much ammonium nitrate is needed to produce the given amount of steam using proportion.

📝 Teacher's Note: Connect this to stoichiometry - the mass ratios are fixed by the balanced equation. Practice with different starting quantities.

🎯 Exam Tip: Set up proportion correctly: known mass/known volume = unknown mass/unknown volume. Cross-multiply to solve.

Question. (iii) 80 parts by weight of \( NH_4NO_3 \) contains 48 parts by weight of oxygen
Answer: So 100 parts will contain = 48 × 100 / 80 = 60 %
In simple words: We calculate what percentage of the compound's weight comes from oxygen atoms.

📝 Teacher's Note: Show students how to count oxygen atoms in the formula (3 oxygen atoms with atomic mass 16 each = 48). This reinforces formula interpretation.

🎯 Exam Tip: For percentage composition, always use the pattern: (mass of element/molecular mass) × 100.

Solution 2007-2

Question. (i) Empirical formula of compound:

Answer: Empirical formula of the compound is \( CBr_3 \).
In simple words: We find the simplest whole number ratio of atoms in the compound using percentage composition.

📝 Teacher's Note: Teach the systematic approach: percentage → moles → ratio → whole numbers. Use the smallest mole value to divide all others.

🎯 Exam Tip: Always divide by the smallest mole value first. If you don't get whole numbers, multiply all ratios by 2 or 3.

Question. (ii) Vapour density = 252
Answer: Empirical formula mass = 12 + 3 × 80 = 252
Molecular mass = 2 × Vapour density
= 2 × 252 = 504
Now, n = Molecular mass / Empirical Formula Mass
= 504/252 = 2
Molecular formula = n × Empirical Formula
= 2 × (\( CBr_3 \))
= \( C_2Br_6 \)
In simple words: The molecular formula is always a whole number multiple of the empirical formula.

📝 Teacher's Note: Emphasize the relationship: Molecular mass = 2 × Vapour density. This is a key formula students must remember.

🎯 Exam Tip: Calculate empirical formula mass first, then use n = molecular mass/empirical mass to find the multiplying factor.

Question. (iii) This substance can be prepared by substitution method.
Answer: This substance can be prepared by substitution method.
In simple words: Carbon tetrabromide can be made by replacing hydrogen atoms in methane with bromine atoms.

📝 Teacher's Note: Explain substitution reactions where hydrogen atoms are replaced by halogens in the presence of sunlight or heat.

🎯 Exam Tip: For halogen compounds of carbon, substitution method is the standard preparation. Mention UV light or heat as conditions.

Solution 2008-1

Question. The gas laws which relates the volume of a gas to the number of molecules of the gas is
Answer: avogadro's law
In simple words: Avogadro's law says equal volumes of gases at the same conditions have equal numbers of molecules.

📝 Teacher's Note: Connect this to the molar volume concept - one mole of any gas occupies 22.4 L at STP, containing Avogadro's number of molecules.

🎯 Exam Tip: Remember: Avogadro's law links volume to number of molecules/moles, not mass or pressure.

Solution 2008-2

Question. (i) \( 2C_8H_{18} + 25O_2 → 16CO_2 + 18H_2O \)
Answer: From the equation:
2 moles of \( C_8H_{18} \) produces = 16 moles of \( CO_2 \)
1 mole of \( C_8H_{18} \) will produce = 16/2 = 8 moles of \( CO_2 \)
So, 8 moles of \( CO_2 \) is produced.
In simple words: We use the balanced equation to find how many moles of carbon dioxide are formed from one mole of octane.

📝 Teacher's Note: Show students how to read mole ratios directly from the coefficients in balanced equations. Practice with different examples.

🎯 Exam Tip: Always write the mole ratio from the equation first, then set up proportion for the given quantity.

Question. (ii) Now, 1 mole of \( CO_2 \) occupies = 22.4 L at STP
Answer: So, 8 moles of \( CO_2 \) will occupy = 22.4 × 8 = 179.2 L at STP
In simple words: We convert moles to volume using the standard molar volume at STP.

📝 Teacher's Note: Reinforce that 22.4 L/mol applies to any gas at STP (0°C, 1 atm). This is a fundamental constant.

🎯 Exam Tip: For gas volume calculations at STP, always multiply moles by 22.4 L/mol.

Question. (iii) Since 2 moles of \( C_8H_{18} \) produces = 16 moles of \( CO_2 \)
Answer: = 16 × 44 = 704
So by burning 2 moles of octane 704 g of \( CO_2 \) is produced.
In simple words: We calculate the total mass of carbon dioxide produced by multiplying moles with molecular mass.

📝 Teacher's Note: Remind students that mass = moles × molecular mass. The molecular mass of CO₂ is 44 g/mol.

🎯 Exam Tip: For mass calculations, always use: mass = number of moles × molecular mass of the compound.

Question. (iv) Molecular formula of octane = \( C_8H_{18} \)
Answer: Its empirical formula will be = \( C_4H_9 \)
In simple words: The empirical formula shows the simplest whole number ratio of atoms in the compound.

📝 Teacher's Note: Show that the empirical formula is found by dividing all subscripts by their highest common factor (2 in this case).

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

Solution 2008-3

Question. (a) (i)

Answer: Empirical formula of the compound is \( CHCl_2 \).
In simple words: The compound has 1 carbon, 1 hydrogen, and 2 chlorine atoms in its simplest form.

📝 Teacher's Note: Use this example to show that percentage composition must always add up to 100%. If not, the remaining percentage is often oxygen.

🎯 Exam Tip: Check your arithmetic - percentages should total 100%, and always divide by the smallest mole value first.

Question. (ii) Now, empirical formula mass = 12 + 1 + 35.5 × 2
Answer: = 12 + 1 + 71 = 84
n = Relative molecular mass/Empirical Formula mass
= 168 / 84 = 2
So, molecular formula = n × (Empirical Formula)
= 2 × \( CHCl_2 \)
= \( C_2H_2Cl_4 \)
In simple words: The actual molecular formula is twice the empirical formula since n equals 2.

📝 Teacher's Note: Emphasize calculating empirical mass carefully before finding n. Common errors occur in this arithmetic step.

🎯 Exam Tip: Double-check your empirical formula mass calculation - add up all atomic masses with their correct subscripts.

Question. (iii) Addition reaction with chlorine.
Answer: Addition reaction with chlorine.
In simple words: This compound can be prepared by adding chlorine molecules to unsaturated hydrocarbons.

📝 Teacher's Note: Explain that addition reactions occur when atoms or groups add to unsaturated compounds without anything being removed.

🎯 Exam Tip: For chlorinated hydrocarbons, mention addition to alkenes or substitution in alkanes as preparation methods.

Question. (b) (i) \( C + 2H_2SO_4 → CO_2 + 2H_2O + 2SO_2 \)
Answer: From the equation:
2 moles of sulphuric acid oxidizes 1 mole of carbon
i.e. 2 × 98 g of sulphuric acid oxidizes 12 g of carbon
So, 49 g of sulphuric acid will oxidize = 12 × 49 / 196 = 3 g
3 g of carbon is oxidized by 49 g of sulphuric acid.
In simple words: We find how much carbon can be oxidized by a given amount of sulphuric acid using proportion.

📝 Teacher's Note: Show students how to extract mass ratios from balanced equations. The coefficients give mole ratios, which convert to mass ratios.

🎯 Exam Tip: Always convert equation coefficients to mass ratios using molecular masses, then apply proportion.

Question. (ii) Again from the equation:
Answer: 2 moles of sulphuric acid liberates 2 moles of sulphur dioxide.
i.e. 196 g of sulphuric acid liberates = 44.8 dm³ of sulphur dioxide
49 g of sulphuric acid will liberate = 44.8 × 49 / 196 = 11.2 dm³
Hence, 11.2 dm³ of sulphur dioxide is liberated at the same time.
In simple words: From the same reaction, we calculate how much sulphur dioxide gas is produced.

📝 Teacher's Note: Connect this to gas stoichiometry - 2 moles of any gas at STP occupy 2 × 22.4 L = 44.8 L.

🎯 Exam Tip: For gas products, convert moles to volume using 22.4 L/mol at STP, then use proportion for the given quantity.

Solution 2009-2

Question. (a) Applying Gay-Lussac's law on the equation:
Answer: \( 2C_2H_2 (g) + 5O_2 (g) → 4CO_2 (g) + 2H_2O (g) \)
2 vol. 5 vol. 4 vol.
As 2 volume of acetylene requires = 5 volume of oxygen
So, 200 cm³ of acetylene will require = 5 × 200 / 2 = 500 cm³
Now further, 2 volume of acetylene produces = 4 volume of carbon dioxide
So, 200 cm³ of acetylene will produce = 4 × 200 / 2 = 400 cm³
Hence, 500 cm³ of oxygen and 400 cm³ of carbon dioxide is formed.
In simple words: Gas volumes in reactions follow simple ratios based on the balanced equation coefficients.

📝 Teacher's Note: Emphasize Gay-Lussac's law - volumes of reacting gases are in simple ratios at same temperature and pressure. Write volume ratios above the equation.

🎯 Exam Tip: For gas volume calculations, extract volume ratios from coefficients, then use proportion. No need to convert to moles first.

Question. (b) Calculation of empirical formula:
Answer: Percentage of hydrogen = 12.5%
Percentage of nitrogen = 100 - 12.5 = 87.5%

Empirical formula = \( NH_2 \)
Given molecular mass = 37
Empirical formula mass = 16
n = Molecular mass / Empirical Formula mass
= 37 / 16 = 2.3 or approximately 2
Molecular formula = n × Empirical formula
= 2 × \( NH_2 \)
= \( N_2H_4 \)
In simple words: This compound (hydrazine) has the molecular formula that is twice its empirical formula.

📝 Teacher's Note: When n is not exactly a whole number, round to the nearest integer. Small experimental errors can cause slight deviations.

🎯 Exam Tip: If the calculated n value is close to a whole number (like 2.3 ≈ 2), round it. Perfect whole numbers are rare in real data.

Solution 2009-3

Question. The correct statement is that equal volumes of all gases under identical conditions contain the same number of molecules.
Answer: The correct statement is that equal volumes of all gases under identical conditions contain the same number of molecules.
In simple words: This is Avogadro's law - any gas at the same temperature and pressure will have the same number of molecules in equal volumes.

📝 Teacher's Note: Connect this to the kinetic theory of gases and explain why this law applies regardless of the gas type - it's about molecule count, not mass.

🎯 Exam Tip: Remember the key phrase "identical conditions" - temperature and pressure must be the same for Avogadro's law to apply.

Solution 2009-4

Question. (a) (i) Molecular weight of nitrogen = 28
Answer: As 6.023 × 10²³ molecules of nitrogen weigh = 28
24 × 10²³ molecules will weigh = 28 × 24 × 10²³ / 6.023 × 10²³
= 28 × 40 = 1120 g
In simple words: We find the mass of a given number of molecules using Avogadro's number as a conversion factor.

📝 Teacher's Note: Emphasize that 6.023 × 10²³ is Avogadro's number - the number of particles in one mole. This is a fundamental constant.

🎯 Exam Tip: Set up proportion: 6.023 × 10²³ molecules : 28g = given molecules : unknown mass. Cross-multiply to solve.

Question. (ii) As 6.023 × 10²³ molecules of nitrogen occupies = 22.4 dm³ at STP
Answer: 24 × 10²³ molecules will occupy = 22.4 × 24 × 10²³ / 6.023 × 10²³
= 896 dm³
In simple words: We calculate the volume occupied by a given number of gas molecules at standard conditions.

📝 Teacher's Note: Reinforce that one mole of any gas occupies 22.4 L at STP. This combines Avogadro's law with the ideal gas law.

🎯 Exam Tip: For gas volume problems, use proportion with Avogadro's number and 22.4 L/mol at STP.

Question. (b) \( NaCl + AgNO_3 → AgCl + NaNO_3 \)
Answer: As 143 g of AgCl is obtained from = 58 g of NaCl
So, 14.3 g of AgCl will be obtained from = 58 × 14.3 / 143 = 5.8 g of NaCl
Weight of commercial NaOH = 30 g
Percentage of NaCl in NaOH = 5.8 × 100 / 30 = 19.33%
In simple words: We find how much pure salt is present in the commercial sample by working backwards from the precipitate formed.

📝 Teacher's Note: This is a typical purity calculation. The precipitate weight tells us about the original pure compound, not the impure sample.

🎯 Exam Tip: For purity problems, find the pure substance amount first, then calculate percentage in the original sample.

Question. (c) As at STP 100 cm³ of gas weighs = 0.5 g
Answer: So, at STP 22400 cm³ of gas will weigh = 0.5 × 22400 / 100 = 112 g
In simple words: We use the density of the gas to find its molecular mass, since 22400 cm³ is the volume of one mole at STP.

📝 Teacher's Note: The molecular mass of a gas equals its mass per mole, which is the mass of 22.4 L at STP. This connects density to molecular mass.

🎯 Exam Tip: For gas density problems, scale up to find the mass of 22.4 L - this gives the molecular mass directly.

Solution 2009-1

Question. The relative molecular mass of the gas is 10.
Answer: The relative molecular mass of the gas is 10.
In simple words: This means the gas molecules are 10 times heavier than hydrogen atoms.

📝 Teacher's Note: Relative molecular mass compares the mass of molecules to hydrogen atoms (mass = 1). It's the same as molecular mass in atomic mass units.

🎯 Exam Tip: Relative molecular mass is numerically equal to molecular mass - no units needed, just the number.