Maharashtra Board Class 11 Chemistry Chapter 1 Some Basic Concepts of Solutions

Chapter 1 Some Basic Concepts of Chemistry

Choose the Most Correct Option

Question A. A sample of pure water, whatever the source always contains ............. by mass of oxygen and 11.1 % by mass of hydrogen.
(a) 88.9
(b) 18
(c) 80
(d) 16
Answer: (a) 88.9
In simple words: Pure water is always made of the same elements in the same ratio by mass. Since hydrogen is 11.1%, oxygen must make up the remaining 88.9% to total 100%.

🎯 Exam Tip: Remember the Law of Definite Proportions: a chemical compound always contains its component elements in a fixed ratio by mass, regardless of its source.

Question B. Which of the following compounds can NOT demonstrate the law of multiple proportions?
(a) NO, NO₂
(b) CO, CO₂
(c) H₂O, H₂O₂
(d) Na₂S, NaF
Answer: (d) Na₂S, NaF
In simple words: The law of multiple proportions applies to pairs of compounds containing the same two elements. Since Na₂S and NaF contain different elements (sulfur and fluorine), they cannot show this law.

🎯 Exam Tip: To quickly identify which pair does not show the law of multiple proportions, look for pairs that do not consist of the exact same two elements.

Question C. Which of the following temperature will read the same value on celsius and Fahrenheit scales.
(a) – 40°
(b) + 40°
(c) – 80°
(d) – 20°
Answer: (a) – 40°
In simple words: At -40 degrees, both the Celsius and Fahrenheit temperature scales show the exact same number.

🎯 Exam Tip: You can prove this mathematically using the conversion formula: \( F = \frac{9}{5}C + 32 \). Substituting \( F = C = x \) gives \( x = -40 \).

Question D. SI unit of the quantity electric current is

Question E. In the reaction \( \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \), the ratio by volume of \( \text{N}_2 \), \( \text{H}_2 \) and \( \text{NH}_3 \) is 1 : 3 : 2. This illustrates the law of
(a) definite proportion
(b) reciprocal proportion
(c) multiple proportion
(d) gaseous volumes
Answer: (d) gaseous volumes
In simple words: When gases react, their volumes combine in simple whole-number ratios, which is the core idea of the law of gaseous volumes.

🎯 Exam Tip: Remember that Gay-Lussac's law of gaseous volumes applies only when all reactants and products are measured at the same temperature and pressure.

Question F. Which of the following has maximum number of molecules?
(a) 7 g \( \text{N}_2 \)
(b) 2 g \( \text{H}_2 \)
(c) 8 g \( \text{O}_2 \)
(d) 20 g \( \text{NO}_2 \)
Answer: (b) 2 g \( \text{H}_2 \)
In simple words: Because hydrogen molecules are extremely light, even a small mass of 2 grams contains a large number of molecules compared to heavier gases of the same or larger weights.

🎯 Exam Tip: To find the maximum number of molecules, calculate the number of moles for each option using \( \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \). The substance with the highest number of moles will always have the most molecules.

Question G. How many g of \( \text{H}_2\text{O} \) are present in 0.25 mol of it?
(a) 4.5
(b) 18
(c) 0.25
(d) 5.4
Answer: (a) 4.5
In simple words: One mole of water weighs 18 grams, so a quarter of a mole (0.25 mol) weighs a quarter of that amount, which is 4.5 grams.

🎯 Exam Tip: Always write down the molar mass of the compound first, then multiply it by the given number of moles to get the correct mass in grams.

Question H. The number of molecules in \( 22.4\text{ cm}^3 \) of nitrogen gas at STP is
(a) \( 6.022 \times 10^{20} \)
(b) \( 6.022 \times 10^{23} \)
(c) \( 22.4 \times 10^{20} \)
(d) \( 22.4 \times 10^{23} \)
Answer: (a) \( 6.022 \times 10^{20} \)
In simple words: Since \( 22400\text{ cm}^3 \) (which is 22.4 liters) of any gas at STP contains \( 6.022 \times 10^{23} \) molecules, a much smaller volume of \( 22.4\text{ cm}^3 \) will contain a thousand times fewer molecules, which is \( 6.022 \times 10^{20} \).

🎯 Exam Tip: Remember that \( 1\text{ dm}^3 = 1000\text{ cm}^3 \). Always convert the given volume to liters or cubic decimeters before using the standard molar volume of \( 22.4\text{ L} \) to avoid calculation errors.

Question I. Which of the following has the largest number of atoms ?
(a) \( 1\text{ g Au}_{(s)} \)
(b) \( 1\text{ g Na}_{(s)} \)
(c) \( 1\text{ g Li}_{(s)} \)
(d) \( 1\text{ g Cl}_{2(g)} \)
Answer: (c) \( 1\text{ g Li}_{(s)} \)
In simple words: Since lithium has the smallest atomic mass among the choices, one gram of lithium contains the highest number of moles, and therefore has the most atoms.

🎯 Exam Tip: For a fixed mass of different elements, the number of atoms is inversely proportional to the atomic mass. The lighter the atom, the more atoms you get in one gram.

Answer the Following Questions

Question A. State and explain Avogadro’s law.
Answer:
(i) In the year 1811, Avogadro made a distinction between atoms and molecules and thereby proposed Avogadro’s law. This law helps in determining the molecular formulae of various gaseous elements and compounds.
(ii) Avogadro proposed that, “Equal volumes of all gases at the same temperature and pressure contain equal number of molecules”.
e.g. Hydrogen gas combines with oxygen gas to produce water vapour as follows:
\( \text{Hydrogen}_{(g)} + \text{Oxygen}_{(g)} \longrightarrow \text{Water}_{(g)} \)
[100 mL] [50 mL] [100 mL]
[2 vol] [1 vol] [2 vol]
According to Avogadro’s law, if 1 volume contains n molecules, then 2n molecules of hydrogen combine with n molecules of oxygen to give 2n molecules of water, i.e., 2 molecules of hydrogen gas combine with 1 molecule of oxygen to give 2
In simple words: Avogadro's law states that if you have equal volumes of different gases under the same temperature and pressure, they will contain the exact same number of molecules. For example, two volumes of hydrogen react with one volume of oxygen to produce two volumes of water vapor.

🎯 Exam Tip: When stating Avogadro's law, always mention the condition of "same temperature and pressure" as these are crucial constant parameters for the law to hold true.

Question B. Point out the difference between \( 12\text{ g} \) of carbon and \( 12\text{ u} \) of carbon.
Answer: \( 12\text{ g} \) of carbon is the molar mass of carbon while \( 12\text{ u} \) of carbon is the mass of one carbon atom. This distinction is fundamental to understanding chemical stoichiometry.
In simple words: \( 12\text{ g} \) is the mass of a huge collection of carbon atoms (one mole), whereas \( 12\text{ u} \) is the mass of just a single carbon atom.

🎯 Exam Tip: Remember that 'g' represents molar mass (grams per mole), while 'u' represents unified atomic mass unit, which is used for single atoms or molecules.

Question C. How many grams does an atom of hydrogen weigh?
Answer: The mass of a hydrogen atom is \( 1.6736 \times 10^{-24}\text{ g} \). This extremely small value reflects the microscopic nature of individual atoms.
In simple words: A single hydrogen atom is incredibly light, weighing only a tiny fraction of a single gram.

🎯 Exam Tip: Always write the unit 'g' and ensure the negative exponent is written clearly to avoid losing marks.

Question D. Calculate the molecular mass of the following in u.
a. \( \text{NH}_3 \)
b. \( \text{CH}_3\text{COOH} \)
c. \( \text{C}_2\text{H}_5\text{OH} \)
Answer:
i. Molecular mass of \( \text{NH}_3 \) = \( (1 \times \text{Average atomic mass of N}) + (3 \times \text{Average atomic mass of H}) \)
= \( (1 \times 14.0\text{ u}) + (3 \times 1.0\text{ u}) \)
= \( 17\text{ u} \)

ii. Molecular mass of \( \text{CH}_3\text{COOH} \) = \( (2 \times \text{Average atomic mass of C}) + (4 \times \text{Average atomic mass of H}) + (2 \times \text{Average atomic mass of O}) \)
= \( (2 \times 12.0\text{ u}) + (4 \times 1.0\text{ u}) + (2 \times 16.0\text{ u}) \)
= \( 60\text{ u} \)

iii. Molecular mass of \( \text{C}_2\text{H}_5\text{OH} \) = \( (2 \times \text{Average atomic mass of C}) + (6 \times \text{Average atomic mass of H}) + (1 \times \text{Average atomic mass of O}) \)
= \( (2 \times 12.0\text{ u}) + (6 \times 1.0\text{ u}) + (1 \times 16.0\text{ u}) \)
= \( 46\text{ u} \)

Ans:
i. The molecular mass of \( \text{NH}_3 \) = \( 17\text{ u} \)
ii. The molecular mass of \( \text{CH}_3\text{COOH} \) = \( 60\text{ u} \)
iii. The molecular mass of \( \text{C}_2\text{H}_5\text{OH} \) = \( 46\text{ u} \). These calculations are based on the standard atomic masses of carbon, hydrogen, nitrogen, and oxygen.
In simple words: To find the molecular mass, we simply add up the atomic masses of all the individual atoms present in the chemical formula.

🎯 Exam Tip: Show each step of the calculation clearly, including the atomic masses of individual elements, to secure full step-wise marks.

Question E. How many particles are present in 1 mole of a substance ?
Answer: The number of particles in one mole is \( 6.0221367 \times 10^{23} \). This extremely large number is universally known as Avogadro's constant.
In simple words: One mole of any substance always contains a fixed, huge number of tiny particles, which is about 602 sextillion particles.

🎯 Exam Tip: Always write the power of 10 correctly as positive 23, not negative, since it represents an incredibly large quantity of particles.

Question F. What is the SI unit of amount of a substance ?
Answer: The SI unit for the amount of a substance is mole (mol). It is one of the seven fundamental base units defined in the International System of Units.
In simple words: The mole is just a standard unit we use to count huge numbers of extremely small things like atoms or molecules.

🎯 Exam Tip: Always write the unit symbol as 'mol' without an 'e' at the end when expressing numerical values in calculations.

Question G. What is meant by molar volume of a gas ?
Answer: The volume occupied by one mole of a gas at standard temperature (\( 0\ ^\circ\text{C} \)) and pressure (\( 1\text{ atm} \)) (STP) is called as molar volume of a gas. The molar volume of a gas at STP is \( 22.4\text{ dm}^3 \). This value remains constant for all ideal gases under these standard conditions.
In simple words: Molar volume is the amount of space that one mole of any gas takes up at standard temperature and pressure, which is always 22.4 liters.

🎯 Exam Tip: Do not forget to mention the specific conditions of temperature (0 °C) and pressure (1 atm) when defining molar volume to secure full marks.

Question H. State and explain the law of conservation of mass.
Answer: Law of conservation of mass:

• The law of conservation of mass states that, “Mass can neither be created nor destroyed” during chemical combination of matter.
• Antoine Lavoisier who is often referred to as the father of modern chemistry performed careful experimental studies for various combustion reactions, namely burning of phosphorus and mercury in the presence of air.
• Both his experiments resulted in increased weight of products.
• After several experiments, in burning of phosphorus, he found that the weight gained by the phosphorus was exactly the same as the weight lost by the air. Hence, \( \text{total mass of reactants} = \text{total mass of products} \).
• When hydrogen gas burns and combines with oxygen to form water, the mass of the water formed is equal to the mass of the hydrogen and oxygen consumed. Thus, this is in accordance with the law of conservation of mass. This fundamental principle ensures that matter is simply rearranged, not lost, during a chemical reaction.

🎯 Exam Tip: To get full marks, state the law clearly in quotes, mention Antoine Lavoisier's name, and write down the key equation: Total mass of reactants = Total mass of products.

Question I. State the law of multiple proportions.
Answer: The law states that, "When two elements A and B form more than one compounds, the masses of element B that combine with a given mass of A are always in the ratio of small whole numbers". This fundamental chemical law helps us understand how elements combine in fixed, predictable proportions to form different substances.
In simple words: If two elements combine to make different compounds, the different weights of one element that combine with a fixed weight of the other will always form simple ratios like 1:2 or 2:3.

🎯 Exam Tip: Be sure to state the law exactly as written in the textbook, highlighting the phrase "ratio of small whole numbers" to secure full marks.

Give One Example of Each

Question A. Homogeneous mixture
Answer: Homogeneous mixture: Solution (An aqueous solution of sugar). In this mixture, the sugar dissolves completely and is distributed uniformly throughout the water.
In simple words: A homogeneous mixture is completely uniform, meaning you cannot see the individual ingredients separately once they are mixed.

🎯 Exam Tip: Sugar dissolved in water or salt dissolved in water are excellent, easy-to-remember examples of homogeneous mixtures.

Question B. Heterogeneous mixture
Answer: Heterogeneous mixture: Suspension (of sand in water). The sand particles do not dissolve and remain suspended, creating a non-uniform composition throughout the mixture.
In simple words: A heterogeneous mixture is not uniform, so you can easily see the different parts mixed together, like sand floating in water.

🎯 Exam Tip: Clearly state that the components of a heterogeneous mixture remain physically separate and visible to the naked eye.

Question C. Element
Answer: Element: Gold. Gold is a pure substance consisting of only one type of atom that cannot be broken down into simpler substances by chemical means.
In simple words: An element is a pure substance made of only one kind of atom, like pure gold, iron, or oxygen.

🎯 Exam Tip: Always choose simple, well-known elements like Gold (Au) or Oxygen (O) to avoid any confusion during evaluation.

Question D. Compound
Answer: Compound: Distilled water. Distilled water is a pure chemical substance made of hydrogen and oxygen atoms chemically bonded together in a fixed 2:1 ratio.
In simple words: A compound is formed when two or more different elements chemically join together in a fixed ratio, like water (\( \text{H}_2\text{O} \)).

🎯 Exam Tip: Distilled water is a perfect example of a compound because it contains only pure \( \text{H}_2\text{O} \) molecules without any dissolved minerals or impurities.

Solve Problems:

Question A. What is the ratio of molecules in 1 mole of \( \text{NH}_3 \) and 1 mole of \( \text{HNO}_3 \).
Answer: According to Avogadro's law, 1 mole of any substance contains Avogadro's number of molecules (\( 6.022 \times 10^{23} \) molecules). Therefore, 1 mole of \( \text{NH}_3 \) contains \( 6.022 \times 10^{23} \) molecules and 1 mole of \( \text{HNO}_3 \) also contains \( 6.022 \times 10^{23} \) molecules.
\( \text{Ratio} = \frac{6.022 \times 10^{23}}{6.022 \times 10^{23}} = 1:1 \). This equal ratio holds true regardless of the chemical formula or molar mass of the compounds involved.
In simple words: One mole of any substance always contains the exact same number of molecules. Since both have one mole, their ratio of molecules is simply 1 to 1.

🎯 Exam Tip: Remember that the number of molecules in a mole is always constant (Avogadro's number), so the ratio of molecules for equal moles of any two substances is always 1:1.

Question B. Calculate number of moles of hydrogen in 0.448 litre of hydrogen gas at STP.
Answer:
Given: Volume of hydrogen at STP = 0.448 L
To find: Number of moles of hydrogen
Formula: Number of moles of a gas \( (n) = \frac{\text{Volume of a gas at STP}}{\text{Molar volume of a gas}} \)
Molar volume of a gas = \( 22.4 \text{ dm}^3\text{ mol}^{-1} = 22.4 \text{ L} \) at STP
Number of moles of a gas \( (n) = \frac{0.448 \text{ L}}{22.4 \text{ L mol}^{-1}} \)

\( \implies \) \( n = 0.02 \text{ mol} \)
Ans: Number of moles of hydrogen = 0.02 mol.
In simple words: To find the number of moles of a gas at standard temperature and pressure (STP), we divide the given volume of the gas by the standard molar volume of 22.4 liters. Dividing 0.448 L by 22.4 L gives us 0.02 moles.

🎯 Exam Tip: Always write down the given values and the formula clearly before doing calculations to ensure you get step-wise marks even if you make a calculation error.

Question C. The mass of an atom of hydrogen is 1.008 u. What is the mass of 18 atoms of hydrogen.
Answer:
Mass of 1 atom of hydrogen = 1.008 u
\( \therefore \) Mass of 18 atoms of hydrogen = \( 18 \times 1.008 \text{ u} = 18.144 \text{ u} \)
Ans: The mass of 18 atoms of hydrogen = 18.144 u.
In simple words: If one single hydrogen atom has a mass of 1.008 u, then 18 hydrogen atoms will weigh 18 times that amount, which equals 18.144 u.

🎯 Exam Tip: Pay close attention to the unit 'u' (unified atomic mass unit) and make sure to include it in your final answer to avoid losing marks.

Question D. Calculate the number of atom in each of the following (Given : Atomic mass of I = 127 u).
a. 254 u of iodine (I)
b. 254 g of iodine (I)
Answer:
a. 254 u of iodine (I):
Let the number of atoms be \( x \).
Atomic mass of iodine (I) = 127 u
\( \times \) Mass of one iodine atom = 127 u
\( \therefore x = \frac{254\text{ u}}{127\text{ u}} = 2 \) atoms

b. 254 g of iodine (I):
Atomic mass of iodine (I) = 127 u
Molar mass of iodine (I) = 127 g/mol
Number of moles of iodine = \( \frac{\text{Mass of iodine}}{\text{Molar mass}} = \frac{254\text{ g}}{127\text{ g/mol}} = 2\text{ moles} \)
Number of atoms = \( \text{Number of moles} \times \text{Avogadro's number} \)

\( \implies \) Number of atoms = \( 2 \times 6.022 \times 10^{23} = 1.2044 \times 10^{24} \) atoms

Ans:
a. 2 atoms of iodine
b. \( 1.2044 \times 10^{24} \) atoms of iodine
In simple words: For part (a), since one iodine atom weighs 127 u, 254 u contains exactly 2 atoms. For part (b), 254 grams of iodine equals 2 moles of iodine, and since one mole contains Avogadro's number of atoms, 2 moles contain \( 1.2044 \times 10^{24} \) atoms.

🎯 Exam Tip: Remember the difference between 'u' (atomic mass unit for individual atoms) and 'g' (grams for molar quantities). Use Avogadro's number only when dealing with grams or moles, not when dealing with atomic mass units (u).

Question. Calculate the number of atoms in 254 g of iodine (I).
Answer:
Atomic mass of iodine = 127 u
\( \therefore \) Molar mass of iodine = \( 127 \text{ g mol}^{-1} \)
Now,
\( \text{Number of moles} = \frac{\text{Mass of a substance}}{\text{Molar mass of a substance}} \)
\( = \frac{254 \text{ g}}{127 \text{ g mol}^{-1}} \)
\( = 2 \text{ mol} \)

Now,
Number of atoms = Number of moles \( \times \) Avogadro’s constant
\( = 2 \text{ mol} \times 6.022 \times 10^{23} \text{ atoms/mol} \)
\( = 12.044 \times 10^{23} \text{ atoms} \)
\( = 1.2044 \times 10^{24} \text{ atoms} \)

Ans.
i. Number of iodine atoms in 254 u = 2 atoms
ii. Number of iodine atoms in 254 g = \( 1.2044 \times 10^{24} \) atoms
In simple words: To find the number of atoms in a given mass, we first calculate the number of moles by dividing the mass by the molar mass, and then multiply those moles by Avogadro's number.

🎯 Exam Tip: Always write down the formula for the number of moles clearly before substituting values to secure step-by-step marks.

Question E. A student used a carbon pencil to write his homework. The mass of this was found to be 5 mg. With the help of this calculate:
a. The number of moles of carbon in his homework writing.
b. The number of carbon atoms in 12 mg of his homework writting.
Answer:
a. For 5 mg of carbon:
\( 5 \text{ mg carbon} = 5 \times 10^{-3} \text{ g carbon} \)
Atomic mass of carbon = 12 u
\( \therefore \) Molar mass of carbon = \( 12 \text{ g mol}^{-1} \)
Now,
\( \text{Number of moles} = \frac{\text{Mass of a substance}}{\text{Molar mass of a substance}} \)
\( = \frac{5 \times 10^{-3} \text{ g}}{12 \text{ g mol}^{-1}} \)
\( = 4.167 \times 10^{-4} \text{ mol} \)

b. For 12 mg of carbon:
\( 12 \text{ mg carbon} = 12 \times 10^{-3} \text{ g carbon} \)
Now,
\( \text{Number of moles} = \frac{\text{Mass of a substance}}{\text{Molar mass of a substance}} = \frac{12 \times 10^{-3} \text{ g}}{12 \text{ g mol}^{-1}} = 1 \times 10^{-3} \text{ mol} \)

Number of atoms = Number of moles \( \times \) Avogadro’s constant
Number of atoms of carbon = \( 1 \times 10^{-3} \text{ mol} \times 6.022 \times 10^{23} \text{ atoms/mol} \)
\( = 6.022 \times 10^{20} \text{ atoms} \)
In simple words: First, convert the mass from milligrams to grams by multiplying by \( 10^{-3} \). Then, find the moles by dividing by carbon's atomic mass (12), and multiply by Avogadro's number to get the total number of atoms.

🎯 Exam Tip: Remember to convert milligrams (mg) to grams (g) by multiplying by \( 10^{-3} \) before performing any mole calculations.

Question F. Arjun purchased 250 g of glucose (\( \text{C}_6\text{H}_{12}\text{O}_6 \)) for Rs. 40. Find the cost of glucose per mole.
Answer:
Given: Mass of glucose = 250 g, cost for 250 g glucose = Rs. 40, molecular formula of glucose = \( \text{C}_6\text{H}_{12}\text{O}_6 \)
To find: Cost per mole of glucose
Calculation: Molecular formula of glucose is \( \text{C}_6\text{H}_{12}\text{O}_6 \).
Molecular mass of glucose
= \( (6 \times \text{Average atomic mass of C}) + (12 \times \text{Average atomic mass of H}) + (6 \times \text{Average atomic mass of O}) \)
= \( (6 \times 12\text{ u}) + (12 \times 1\text{ u}) + (6 \times 16\text{ u}) \)
= \( 180\text{ u} \)
\( \therefore \) Molar mass of glucose = \( 180\text{ g mol}^{-1} \)

\( \text{Number of moles} = \frac{\text{Mass of a substance}}{\text{Molar mass of a substance}} = \frac{250\text{ g}}{180\text{ g mol}^{-1}} = \frac{250}{180}\text{ mol} \)

Now,
\( \frac{250}{180}\text{ mol} \) of glucose cost = Rs. 40
1 mol glucose cost = \( x \)
\( \therefore x = \frac{40 \times 180}{250} = \text{Rs. } 28.8\text{/mol of glucose} \)
Ans. The cost of glucose per mole is Rs. 28.8. This calculation helps us understand the economic value of chemical substances on a molecular scale.

[ Calculation using log table:
\( \frac{40 \times 180}{250} \)
= \( \text{Antilog}_{10} [\log_{10}(40) + \log_{10}(180) - \log_{10}(250)] \)
= \( \text{Antilog}_{10} [1.6021 + 2.2553 - 2.3979] \)
= \( \text{Antilog}_{10} [1.4595] = 28.80 \) ]
In simple words: We first find the weight of one mole of glucose, which is 180 grams. Then, we use a simple ratio to find how much 180 grams costs if 250 grams costs Rs. 40.

🎯 Exam Tip: Always write down the formula for molecular mass and molar mass clearly before doing calculations to secure step-wise marks.

Question G. The natural isotopic abundance of \( ^{10}\text{B} \) is 19.60% and \( ^{11}\text{B} \) is 80.40 %. The exact isotopic masses are 10.13 and 11.009 respectively. Calculate the average atomic mass of boron.
Answer:
Average atomic mass of Boron (B)
= \( \frac{(\text{Isotopic mass of }^{10}\text{B} \times \% \text{ abundance}) + (\text{Isotopic mass of }^{11}\text{B} \times \% \text{ abundance})}{100} \)
= \( \frac{(10.13 \times 19.60) + (11.009 \times 80.40)}{100} \)
= \( \frac{198.548 + 885.1236}{100} \)
= \( \frac{1083.6716}{100} \)
= \( 10.8367\text{ u} \) (or approx \( 10.84\text{ u} \))
Ans. The average atomic mass of boron is \( 10.84\text{ u} \). This weighted average reflects the relative contribution of each isotope as they occur naturally on Earth.
In simple words: To find the average mass, we multiply each isotope's mass by its percentage in nature, add them together, and divide by 100.

🎯 Exam Tip: Remember that average atomic mass will always lie between the masses of the individual isotopes, closer to the one with the higher abundance.

Question H. Convert the following degree Celsius temperature to degree Fahrenheit.
(a) 40 °C
(b) 30 °C
Answer:
(a) For 40 °C:
Given: Temperature in degree Celsius (\(\text{°C}\)) = 40 °C
To find: Temperature in degree Fahrenheit (\(\text{°F}\))
Formula: \( \text{°F} = \frac{9}{5}(\text{°C}) + 32 \)
Calculation: Substituting 40 °C in the formula,
\( \text{°F} = \frac{9}{5}(40) + 32 \)
\( \implies \text{°F} = 72 + 32 \)
\( \implies \text{°F} = 104\text{ °F} \)

(b) For 30 °C:
Given: Temperature in degree Celsius (\(\text{°C}\)) = 30 °C
To find: Temperature in degree Fahrenheit (\(\text{°F}\))
Formula: \( \text{°F} = \frac{9}{5}(\text{°C}) + 32 \)
Calculation: Substituting 30 °C in the formula,
\( \text{°F} = \frac{9}{5}(30) + 32 \)
\( \implies \text{°F} = 54 + 32 \)
\( \implies \text{°F} = 86\text{ °F} \)

Thus, the temperature of 40 °C corresponds to 104 °F, and 30 °C corresponds to 86 °F. Knowing how to convert between these scales is highly useful for scientific measurements and understanding global weather reports.
In simple words: To convert Celsius to Fahrenheit, multiply the Celsius value by 9, divide it by 5, and then add 32 to the result.

🎯 Exam Tip: Always write down the formula clearly before substituting values to secure step-by-step marks in calculations.

Question I. Calculate the number of moles and molecules of acetic acid present in 22 g of it.
Answer:
Given: Mass of acetic acid = 22 g
To find: The number of moles and molecules of acetic acid
Formula:
1. \( \text{Molar mass of acetic acid (CH}_3\text{COOH)} = (2 \times \text{Atomic mass of C}) + (4 \times \text{Atomic mass of H}) + (2 \times \text{Atomic mass of O}) \)
\( \implies \text{Molar mass} = (2 \times 12) + (4 \times 1) + (2 \times 16) = 24 + 4 + 32 = 60\text{ g/mol} \)
2. \( \text{Number of moles} = \frac{\text{Mass of substance in grams}}{\text{Molar mass of substance}} \)
3. \( \text{Number of molecules} = \text{Number of moles} \times \text{Avogadro's number } (6.022 \times 10^{23}) \)

Calculation:
\( \text{Number of moles} = \frac{22}{60} \approx 0.367\text{ moles} \)
\( \text{Number of molecules} = 0.367 \times 6.022 \times 10^{23} \approx 2.21 \times 10^{23}\text{ molecules} \)

Hence, 22 g of acetic acid contains approximately 0.367 moles and \(2.21 \times 10^{23}\) molecules. This calculation is fundamental to understanding stoichiometry in chemical reactions.
In simple words: First, find the weight of one mole of acetic acid (60 grams). Then, divide the given weight (22 grams) by 60 to get the moles, and multiply that by Avogadro's number to find the total number of molecules.

🎯 Exam Tip: Remember to calculate the molar mass of the compound accurately first, as any error here will make all subsequent steps incorrect.

Question J. 24 g of carbon reacts with some oxygen to make 88 grams of carbon dioxide. Find out how much oxygen must have been used.
Answer:
Given: Mass of carbon (reactant) = 24 g, mass of carbon dioxide (product) = 88 g
To find: Mass of oxygen (reactant)
Calculation: 12 g of carbon combine with 32 g oxygen to form 44 g of carbon dioxide as follows:
\( \text{Carbon} + \text{Oxygen} \longrightarrow \text{Carbon dioxide} \)
\( 12\text{ g} + 32\text{ g} \longrightarrow 44\text{ g} \)
Hence, \( (2 \times 12 = 24\text{ g}) \) of carbon will combine with \( (2 \times 32 = 64\text{ g}) \) of oxygen to give \( (2 \times 44 = 88\text{ g}) \) carbon dioxide. This calculation perfectly demonstrates the law of conservation of mass where the total mass of reactants equals the total mass of products.
Ans: Mass of oxygen used = 64 g
In simple words: Since the total weight before and after a chemical reaction must stay the same, we subtract the 24 grams of carbon from the 88 grams of carbon dioxide to find that 64 grams of oxygen were used.

🎯 Exam Tip: Always state the law of conservation of mass when solving mass-relation problems to show the examiner the scientific principle behind your calculation.

Question K. Calculate number of atoms in each of the following. (Average atomic mass : N = 14 u, S = 32 u)
a. 0.4 mole of nitrogen
b. 1.6 g of sulfur
Answer:
a. 0.4 mole of nitrogen (N):
Number of moles of nitrogen atoms = 0.4 mol
Number of atoms = Number of moles \( \times \) Avogadro's constant
\( = 0.4 \times 6.022 \times 10^{23} \)
\( = 2.409 \times 10^{23} \) atoms of nitrogen.

b. 1.6 g of sulfur (S):
Number of moles of sulfur = \( \frac{\text{Mass of substance}}{\text{Molar mass of sulfur}} \)
\( = \frac{1.6\text{ g}}{32\text{ g/mol}} = 0.05\text{ mol} \)
Number of atoms = Number of moles \( \times \) Avogadro's constant
\( = 0.05 \times 6.022 \times 10^{23} \)
\( = 3.011 \times 10^{22} \) atoms of sulfur.
In simple words: To find the number of atoms, we multiply the number of moles by Avogadro's number (\( 6.022 \times 10^{23} \)). If we are given grams instead of moles, we first divide the grams by the atomic weight to find the moles.

🎯 Exam Tip: Remember to convert mass into moles first before multiplying by Avogadro's number, and always write down the formula you are using.

Question L. 2.0 g of a metal burnt in oxygen gave 3.2 g of its oxide. 1.42 g of the same metal heated in steam gave 2.27 g of its oxide. Which law is verified by these data?
Answer: Here, metal oxide is obtained by two different methods; reactions of metal with oxygen and reaction of metal with water vapour (steam).

In the first reaction (reaction with oxygen):
The mass of oxygen in metal oxide = \( 3.2\text{ g} - 2.0\text{ g} = 1.2\text{ g} \)
% of oxygen = \( \frac{1.2}{3.2} \times 100 = 37.5\% \)
% of metal = \( \frac{2.0}{3.2} \times 100 = 62.5\% \)

In the second reaction (reaction with steam):
The mass of oxygen in metal oxide = \( 2.27\text{ g} - 1.42\text{ g} = 0.85\text{ g} \)
% of oxygen = \( \frac{0.85}{2.27} \times 100 = 37.44\% \approx 37.5\% \)
% of metal = \( \frac{1.42}{2.27} \times 100 = 62.56\% \approx 62.5\% \)

Therefore, irrespective of the source, the given compound contains the same elements in the same proportion. This consistency shows that chemical compounds are highly stable and uniform in their makeup. The law of definite proportions states that "A given compound always contains exactly the same proportion of elements by weight". Hence, the law of definite proportions is verified by these data.

Ans: The law of definite proportions is verified by given data.
In simple words: No matter how we prepare a chemical compound, it will always contain the exact same elements in the exact same weight ratio. Since both methods gave the same percentage of metal and oxygen, it proves the law of definite proportions.

🎯 Exam Tip: Clearly show the percentage calculations for both reactions. Highlighting that the final percentages are approximately equal is the key to scoring full marks.

Question M. In two moles of acetaldehyde (\( \text{CH}_3\text{CHO} \)) calculate the following:
a. Number of moles of carbon
b. Number of moles of hydrogen
c. Number of moles of oxygen
d. Number of molecules of acetaldehyde
Answer: Molecular formula of acetaldehyde: \( \text{C}_2\text{H}_4\text{O} \)
Moles of acetaldehyde = \( 2 \text{ mol} \)
a. Number of moles of carbon atoms = Moles of acetaldehyde \( \times \) Number of carbon atoms in one molecule
\( = 2 \times 2 \)
\( = 4 \text{ moles of carbon atoms} \)
b. Number of moles of hydrogen atoms = Moles of acetaldehyde \( \times \) Number of hydrogen atoms in one molecule
\( = 2 \times 4 \)
\( = 8 \text{ moles of hydrogen atoms} \)
c. Number of moles of oxygen atoms = Moles of acetaldehyde \( \times \) Number of oxygen atoms in one molecule
\( = 2 \times 1 \)
\( = 2 \text{ moles of oxygen atoms} \)
d. Number of molecules of acetaldehyde = Moles of acetaldehyde \( \times \) Avogadro's number
\( = 2 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol} \)
\( = 12.044 \times 10^{23} \text{ molecules of acetaldehyde} \)
Thus, the number of moles of carbon, hydrogen, and oxygen are 4, 8, and 2 respectively, and the total number of acetaldehyde molecules is \( 12.044 \times 10^{23} \). This calculation clearly demonstrates how molecular formulas relate directly to molar quantities.
In simple words: One molecule of acetaldehyde has 2 carbon, 4 hydrogen, and 1 oxygen atoms. If we have 2 moles of the whole compound, we simply multiply these numbers by 2 to find the moles of each element, and multiply the total moles by Avogadro's number to find the total molecules.

🎯 Exam Tip: Always write down the simplified molecular formula first to avoid counting atoms incorrectly from the structural formula.

Question N. Calculate the number of moles of magnesium oxide, \( \text{MgO} \) in:
i. 80 g and
ii. 10 g of the compound.
(Average atomic masses of \( \text{Mg} = 24 \) and \( \text{O} = 16 \))
Answer: Given:
i. Mass of \( \text{MgO} = 80 \text{ g} \)
ii. Mass of \( \text{MgO} = 10 \text{ g} \)
First, calculate the molar mass of \( \text{MgO} \):
Molar mass of \( \text{MgO} = \text{Atomic mass of Mg} + \text{Atomic mass of O} \)
\( = 24 + 16 = 40 \text{ g/mol} \)
Now, use the formula: \( \text{Number of moles} = \frac{\text{Mass of compound}}{\text{Molar mass of compound}} \)
i. For 80 g of \( \text{MgO} \):
\( \text{Number of moles} = \frac{80 \text{ g}}{40 \text{ g/mol}} = 2 \text{ moles} \)
ii. For 10 g of \( \text{MgO} \):
\( \text{Number of moles} = \frac{10 \text{ g}}{40 \text{ g/mol}} = 0.25 \text{ moles} \)
Therefore, there are 2 moles of magnesium oxide in 80 g, and 0.25 moles in 10 g of the compound. Understanding molar mass is essential for performing stoichiometric calculations in chemical reactions.
In simple words: To find the number of moles, we first find the weight of one mole of MgO, which is 40 grams. Then we divide our given weights (80 grams and 10 grams) by 40 to see how many moles we have.

🎯 Exam Tip: Remember to always state the formula \( \text{Number of moles} = \frac{\text{Mass}}{\text{Molar mass}} \) clearly before substituting values to secure step-wise marks.

Question 1. Calculate the number of moles in:
(i) 80 g of MgO
(ii) 10 g of MgO
Answer:
To find: Number of moles of MgO
Formula: \( \text{Number of moles (n)} = \frac{\text{Mass of a substance}}{\text{Molar mass of a substance}} \)
Calculation:
First, we find the molecular mass of MgO:
\( \text{Molecular mass of MgO} = (1 \times \text{Average atomic mass of Mg}) + (1 \times \text{Average atomic mass of O}) \)
\( = (1 \times 24\text{ u}) + (1 \times 16\text{ u}) = 40\text{ u} \)
\( \therefore \text{Molar mass of MgO} = 40\text{ g mol}^{-1} \)

(i) For 80 g of MgO:
Given: Mass of MgO = 80 g
\( \text{Number of moles (n)} = \frac{80\text{ g}}{40\text{ g mol}^{-1}} = 2\text{ mol} \)

(ii) For 10 g of MgO:
Given: Mass of MgO = 10 g
\( \text{Number of moles (n)} = \frac{10\text{ g}}{40\text{ g mol}^{-1}} = 0.25\text{ mol} \)

This calculation helps us understand the quantitative relationship between mass and chemical amount in a laboratory sample.
Ans:
(i) The number of moles in 80 g of magnesium oxide, MgO = 2 mol
(ii) The number of moles in 10 g of magnesium oxide, MgO = 0.25 mol
In simple words: To find the number of moles, we divide the given mass of magnesium oxide by its molar mass (40 g/mol). So, 80 g gives 2 moles, and 10 g gives 0.25 moles.

🎯 Exam Tip: Always remember to calculate the molar mass first by adding the atomic masses of individual elements, and write the correct unit (g mol⁻¹) for molar mass.

Question O. What is volume of carbon dioxide, \( \text{CO}_2 \) occupying by
(i) 5 moles and
(ii) 0.5 mole
of \( \text{CO}_2 \) gas measured at STP.
Answer:
Given:
(i) Number of moles of \( \text{CO}_2 \) = 5 mol
(ii) Number of moles of \( \text{CO}_2 \) = 0.5 mol
To find: Volume at STP
Formula: \( \text{Number of moles of a gas (n)} = \frac{\text{Volume of a gas at STP}}{\text{Molar volume of a gas}} \)
Calculation:
Molar volume of any gas at STP is \( 22.4\text{ dm}^3\text{ mol}^{-1} \).
\( \text{Number of moles of a gas (n)} = \frac{\text{Volume of a gas at STP}}{\text{Molar volume of a gas}} \)
\( \implies \text{Volume of the gas at STP} = \text{Number of moles of a gas (n)} \times \text{Molar volume of a gas} \)

(i) For 5 moles of \( \text{CO}_2 \):
\( \text{Volume of the gas at STP} = 5\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 112\text{ dm}^3 \)

(ii) For 0.5 mole of \( \text{CO}_2 \):
\( \text{Volume of the gas at STP} = 0.5\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 11.2\text{ dm}^3 \)

This relationship is based on Avogadro's law, which states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.
Ans:
(i) Volume of 5 mol of \( \text{CO}_2 \) = \( 112\text{ dm}^3 \)
(ii) Volume of 0.5 mol of \( \text{CO}_2 \) = \( 11.2\text{ dm}^3 \)
In simple words: One mole of any gas at STP occupies a fixed volume of 22.4 dm³. To find the volume of any number of moles, we simply multiply the number of moles by 22.4.

🎯 Exam Tip: Remember that the molar volume of any ideal gas at STP is always a constant value of 22.4 dm³ (or liters). Clearly state this standard value before starting your calculations.

Question P. Calculate the mass of potassium chlorate required to liberate \(6.72\text{ dm}^3\) of oxygen at STP. Molar mass of \(\text{KClO}_3\) is \(122.5\text{ g mol}^{-1}\).
Answer:
The molecular formula of potassium chlorate is \(\text{KClO}_3\).
Required chemical equation:
\(2\text{KClO}_3 \longrightarrow 2\text{KCl} + 3\text{O}_2 \uparrow\)
[2 moles] \(\quad\quad\quad\quad\quad\) [3 moles]
\(2\text{ moles of KClO}_3 = 2 \times 122.5 = 245\text{ g}\)
\(3\text{ moles of O}_2\text{ at STP occupy} = (3 \times 22.4\text{ dm}^3) = 67.2\text{ dm}^3\)
Thus, \(245\text{ g}\) of potassium chlorate will liberate \(67.2\text{ dm}^3\) of oxygen gas. This stoichiometric relationship allows us to easily scale the reaction for any desired volume of gas.
Let 'x' gram of \(\text{KClO}_3\) liberate \(6.72\text{ dm}^3\) of oxygen gas at S.T.P.
\( \therefore x = \frac{245 \times 6.72}{67.2} = 24.5\text{ g}\)
Ans: Mass of potassium chlorate required = \(24.5\text{ g}\)
In simple words: To find how much starting material we need, we use the balanced chemical equation to compare the mass of potassium chlorate to the volume of oxygen gas it produces at standard conditions. By setting up a simple ratio, we find that we need exactly 24.5 grams.

🎯 Exam Tip: Always write the balanced chemical equation first and clearly state the molar volume of gases at STP (\(22.4\text{ dm}^3\)) to secure full marks.

Question Q. Calculate the number of atoms of hydrogen present in \(5.6\text{ g}\) of urea, \(\text{(NH}_2\text{)}_2\text{CO}\). Also calculate the number of atoms of N, C and O.
Answer:
Given: Mass of urea = \(5.6\text{ g}\)
To find: The number of atoms of hydrogen, nitrogen, carbon and oxygen
Calculation: Molecular formula of urea: \(\text{(NH}_2\text{)}_2\text{CO}\)
Molar mass of urea = \(60\text{ g mol}^{-1}\)
\(\text{Number of moles} = \frac{\text{Mass of a substance}}{\text{Molar mass of a substance}} = \frac{5.6\text{ g}}{60\text{ g mol}^{-1}} = 0.0933\text{ mol}\)
\( \therefore \text{Moles of urea} = 0.0933\text{ mol}\)
\(\text{Number of atoms} = \text{Number of moles} \times \text{Avogadro's constant}\)
Now, 1 molecule of urea has total 8 atoms, out of which 4 atoms are of H, 2 atoms are of N, 1 of C and 1 of O. This atomic composition is key to determining the individual elemental counts.
\( \therefore \text{Number of H atoms in } 5.6\text{ g of urea} = (4 \times 0.0933)\text{ mol} \times 6.022 \times 10^{23}\text{ atoms/mol} = 2.247 \times 10^{23}\text{ atoms of hydrogen}\)
\( \dots \text{Number of N atoms in } 5.6\text{ g of urea} = (2 \times 0.0933)\text{ mol} \times 6.022 \times 10^{23}\text{ atoms/mol} = 1.124 \times 10^{23}\text{ atoms of nitrogen}\)
\( \therefore \text{Number of C atoms in } 5.6\text{ g of urea} = (1 \times 0.0933)\text{ mol} \times 6.022 \times 10^{23}\text{ atoms/mol} = 5.618 \times 10^{22}\text{ atoms of carbon}\)
\( \dots \text{Number of O atoms in } 5.6\text{ g of urea} = (1 \times 0.0933)\text{ mol} \times 6.022 \times 10^{23}\text{ atoms/mol} = 5.618 \times 10^{22}\text{ atoms of oxygen}\)
In simple words: First, we find how many moles of urea are in 5.6 grams by dividing by its molar mass. Then, we multiply the moles by Avogadro's number and the number of each type of atom in one molecule of urea to get the total count of each atom.

🎯 Exam Tip: Remember to multiply the total moles of the compound by the number of atoms of that specific element present in one molecule before multiplying by Avogadro's number.

Question R. Calculate the mass of sulfur dioxide produced by burning \( 16\text{ g} \) of sulfur in excess of oxygen in contact process. (Average atomic mass : S = 32 u, O = 16 u)
Answer:
Given: Mass of sulphur (reactant) = \( 16\text{ g} \)
To find: Mass of sulphur dioxide (product)
Calculation: \( 32\text{ g} \) of sulphur combine with \( 32\text{ g} \) oxygen to form \( 64\text{ g} \) of sulphur dioxide as follows:
\( \text{Sulphur} + \text{Oxygen} \longrightarrow \text{Sulphur dioxide} \)
\( 32\text{ g} \quad + \quad 32\text{ g} \quad \longrightarrow \quad 64\text{ g} \)
Hence, \( (0.5 \times 32 = 16\text{ g}) \) of sulphur will combine with \( (0.5 \times 32 = 16\text{ g}) \) of oxygen to give \( (0.5 \times 64 = 32\text{ g}) \) sulphur dioxide. This stoichiometric calculation is based on the law of conservation of mass.
Ans: Mass of sulphur dioxide produced = \( 32\text{ g} \)
In simple words: When we burn 32 grams of sulfur, we get 64 grams of sulfur dioxide. Since we are only burning half that amount (16 grams), we will get exactly half the amount of sulfur dioxide, which is 32 grams.

🎯 Exam Tip: Always write the balanced chemical equation first to establish the correct stoichiometric ratio between reactants and products.

5. Explain

Question A. The need of the term average atomic mass.
Answer:
• Several naturally occurring elements exist as a mixture of two or more isotopes.
• Isotopes have different atomic masses.
• The atomic mass of such an element is the average of atomic masses of its isotopes.
• For this purpose, the atomic masses of isotopes and their relative percentage abundances are considered.
Hence, the term average atomic mass is needed to express atomic mass of elements containing mixture of two or more isotopes. This ensures that chemical calculations reflect the actual isotopic composition found in nature.
In simple words: Most elements in nature are made of different versions called isotopes, which have different weights. We use average atomic mass to get a single, accurate weight that represents this natural mixture.

🎯 Exam Tip: Remember to mention both 'isotopes' and 'relative percentage abundance' as these are the key terms examiners look for.

Question B. Molar mass.
Answer:
(i) The mass of one mole of a substance (element/compound) in grams is called its molar mass.
(ii) The molar mass of any element in grams is numerically equal to atomic mass of that element in u.
For example:

🎯 Exam Tip: Remember that while atomic mass is expressed in unified mass units (u), molar mass is always expressed in grams per mole (g/mol) even though their numerical values are identical.

Question C. Mole concept.
Answer:
• Even a small amount of any substance contains very large number of atoms or molecules. Therefore, a quantitative adjective ‘mole’ is used to express the large number of sub-microscopic entities like atoms, molecules, ions, electrons, etc. present in a substance.
• Thus, one mole is the amount of a substance that contains as many entities or particles as there are atoms in exactly 12 g (or 0.012 kg) of the carbon-12 isotope. This standard reference helps scientists count extremely tiny particles by weighing them.
In simple words: A mole is a unit used to count huge numbers of tiny particles like atoms or molecules, just like we use the word 'dozen' to mean 12 items. One mole always contains a fixed, massive number of these particles.

🎯 Exam Tip: Always define a mole with reference to the carbon-12 isotope, as this is the standard definition examiners look for to award full marks.

Question D. Formula mass with an example.
Answer: The formula mass of a substance is the sum of atomic masses of the atoms present in the formula. In substances such as sodium chloride, positive sodium (\( \text{Na}^+ \)) and negative chloride (\( \text{Cl}^- \)) entities are arranged in a three-dimensional structure in a way that one sodium ion is surrounded by six chloride ions, all at the same distance from it and vice versa. Thus, sodium chloride does not contain discrete molecules as the constituent units. Therefore, \( \text{NaCl} \) is just the formula that is used to represent sodium chloride though it is not a molecule. In such compounds, the formula (i.e., \( \text{NaCl} \)) is used to calculate the formula mass instead of molecular mass.
e.g. Formula mass of sodium chloride = atomic mass of sodium + atomic mass of chlorine
\( = 23.0\text{ u} + 35.5\text{ u} = 58.5\text{ u} \)
In simple words: Formula mass is the total weight of all atoms in a chemical formula, used especially for ionic compounds like salt that do not form separate molecules.

🎯 Exam Tip: Always remember to use the term 'formula mass' instead of 'molecular mass' for ionic compounds like NaCl, as they exist as crystal lattices rather than individual molecules.

Question E. Molar volume of gas.
Answer:
(i) It is more convenient to measure the volume rather than mass of the gas.
(ii) It is found from Avogadro's law that one mole of any gas occupies a volume of \( 22.4\text{ dm}^3 \) at standard temperature (\( 0\text{ }^\circ\text{C} \)) and pressure (\( 1\text{ atm} \)) (STP).
(iii) The volume of \( 22.4\text{ dm}^3 \) at STP is known as molar volume of a gas.
(iv) The relationship between number of moles and molar volume can be expressed as follows: \[ \text{Number of moles of a gas (n)} = \frac{\text{Volume of the gas at STP}}{\text{Molar volume of the gas}} \] \[ = \frac{\text{Volume of the gas at STP}}{22.4\text{ dm}^3\text{ mol}^{-1}} \]
In simple words: Molar volume is the fixed volume of 22.4 liters that one mole of any gas occupies when kept at standard temperature and pressure.

🎯 Exam Tip: Be sure to memorize the value \( 22.4\text{ dm}^3 \) (or liters) at STP, as it is a key constant used in stoichiometry calculations for gases.

[Note: IUPAC has recently changed the standard pressure to 1 bar. Under these new STP conditions the molar volume of a gas is 22.71 \( \text{L mol}^{-1} \)]

Question F. Types of matter (on the basis of chemical composition).
Answer: Matter on the basis of chemical composition can be classified as follows:
i. Pure substances: They always have a definite chemical composition. They always have the same properties regardless of their origin. This classification helps scientists understand how different materials interact and react with one another. e.g. Pure metal, distilled water, etc.
They are of two types:
a. Elements: They are pure substances, which cannot be broken down into simpler substances by ordinary chemical changes. Elements are further classified into three types:
1. Metals:
• They have a lustre (a shiny appearance).
• They conduct heat and electricity.
• They can be drawn into wire (ductile).
• They can be hammered into thin sheets (malleable).
e.g. Gold, silver, copper, iron. Mercury is a liquid metal at room temperature.
2. Nonmetals:
• They have no lustre, (except diamond, iodine)
• They are poor conductors of heat and electricity, (except graphite)
• They cannot be hammered into sheets or drawn into wire, because they are brittle. e.g. Iodine
3. Metalloids: Some elements have properties that are intermediate between metals and nonmetals and are called metalloids or semimetals.
e.g. Arsenic, silicon and germanium.
b. Compounds: They are the pure substances which are made up of two or more elements in fixed proportion.
e.g. Water, ammonia, methane, etc.
In simple words: On a chemical level, matter is split into pure substances, which include elements (basic building blocks like metals, nonmetals, and metalloids) and compounds (combinations of elements in fixed ratios, like water).

🎯 Exam Tip: When defining elements, always list the three sub-types (metals, nonmetals, and metalloids) along with unique exceptions like mercury and graphite to secure full marks.

Mixtures

They have no definite chemical composition and hence no definite properties. They can be separated by physical methods. For example, paint is a mixture of oils, pigment, and additives, while concrete is a mixture of sand, cement, and water.

Types of Mixtures:

• Homogeneous mixture: In a homogeneous mixture, constituents remain uniformly mixed throughout its bulk (e.g., a solution in which solute and solvent molecules are uniformly mixed throughout).
• Heterogeneous mixture: In a heterogeneous mixture, constituents are not uniformly mixed throughout its bulk (e.g., a suspension which contains insoluble solid in a liquid).

Can You Tell? (Textbook Page No. 1)

Question 1. Which are mixtures and pure substances from the following?
(i) Sea water
(ii) Gasoline
(iii) Skin
(iv) A rusty nail
(v) A page of textbook
(vi) Diamond
Answer:

🎯 Exam Tip: To easily distinguish them, ask yourself if the substance can be separated into different components by simple physical means; if yes, it is a mixture.

Can You Tell? (Textbook Page No. 2)

Question 1. Classify the following as element and compound.
(i) Mercuric oxide
(ii) Helium gas
(iii) Water
(iv) Table salt
(v) Iodine
(vi) Mercury
(vii) Oxygen
(viii) Nitrogen
Answer: This classification helps us understand the fundamental chemical nature of these common substances.

🎯 Exam Tip: Remember that elements consist of only one type of atom and cannot be broken down, while compounds contain different atoms chemically bonded together.

Can You Tell? (Textbook Page No. 6)

Question 1. If 10 volumes of dihydrogen gas react with 5 volumes of dioxygen gas, how many volumes of water vapour would be produced?
Answer:
\( 2\text{H}_{2(g)} + \text{O}_{2(g)} \longrightarrow 2\text{H}_2\text{O}_{(g)} \)
[2 vol]      [1 vol]            [2 vol]

If 10 volumes of dihydrogen gas react with 5 volumes of dioxygen gas, then 10 volumes of water vapour would be produced. This reaction perfectly demonstrates Gay-Lussac's law of combining volumes of gases.
In simple words: According to the balanced chemical equation, two volumes of hydrogen react with one volume of oxygen to produce two volumes of water vapor. Therefore, 10 volumes of hydrogen will react with 5 volumes of oxygen to give exactly 10 volumes of water vapor.

🎯 Exam Tip: Always write the balanced chemical equation first to easily determine the volume ratio of the reacting gases.

Can You Recall? (Textbook Page No. 6)

Question 1. What is an atom and molecule? What is the order of magnitude of mass of one atom? What are isotopes?
Answer:
• The smallest indivisible particle of an element is called an atom.
• A molecule is an aggregate of two or more atoms of definite composition which are held together by chemical bonds.
• Every atom of an element has definite mass. The order of magnitude of mass of one atom is \( 10^{-27} \text{ kg} \).
• Isotopes are the atoms of the same element having same atomic number but different mass number. These different forms of an element share identical chemical properties but differ in their physical properties due to varying neutron counts.
In simple words: Atoms are the basic building blocks of matter, while molecules are groups of atoms bonded together. Atoms are extremely light, weighing around \( 10^{-27} \text{ kg} \), and isotopes are just different versions of the same element with different weights.

🎯 Exam Tip: Clearly define each term in separate bullet points to make your answer easy for the examiner to read and grade.

Try This (Textbook Page No. 8)

Question 1. Find the formula mass of \( \text{CaSO}_4 \), if atomic mass of Ca = 40.1 u, S = 32.1 u and O = 16.0 u.
Answer:
Formula mass of \( \text{CaSO}_4 \)
= Average atomic mass of Ca + Average atomic mass of S + Average atomic mass of four O
= \( (40.1) + 32.1 + (4 \times 16.0) \)
= \( 136.2 \text{ u} \)

Ans: Formula mass of \( \text{CaSO}_4 \) = \( 136.2 \text{ u} \). This calculated value represents the sum of the atomic masses of all atoms present in one formula unit of the ionic compound.
In simple words: To find the formula mass, we simply add up the individual masses of one calcium atom, one sulfur atom, and four oxygen atoms. The total sum gives us the formula mass of the compound.

🎯 Exam Tip: Do not forget to write the unit 'u' (unified mass) at the end of your final answer to avoid losing marks.

Can You Recall? (Textbook Page No. 8)

Question 1.
i. One dozen means how many items?
ii. One gross means how many items?
Answer:
i. One dozen means 12 items.
ii. One gross means 144 items. These standard units of measurement help us easily group and count large quantities of objects in daily life.
In simple words: A dozen is a group of 12 things (like eggs), and a gross is a larger group containing 144 things (which is equal to 12 dozen).

🎯 Exam Tip: Remember that a gross is simply a dozen of dozens (\( 12 \times 12 = 144 \)), which makes it easy to recall during exams.

Try This (Textbook Page No. 10)

Question 1. Calculate the volume in \( \text{dm}^3 \) occupied by \( 60.0\text{ g} \) of ethane at STP.
Answer:
Given: Mass of ethane at STP = \( 60.0\text{ g} \)
To find: Volume of ethane
Formulae:
i. \( \text{Number of moles} = \frac{\text{Mass of a substance}}{\text{Molar mass of the substance}} \)
ii. \( \text{Number of moles} = \frac{\text{Volume of a gas at STP}}{\text{Molar volume of a gas}} \)

Calculation:
Molar volume of a gas = \( 22.4\text{ dm}^3\text{ mol}^{-1} \) at STP
Molecular mass of ethane (\( \text{C}_2\text{H}_6 \)) = \( 30\text{ g mol}^{-1} \)
Using formula (i):
\( \text{Number of moles} = \frac{60.0\text{ g}}{30\text{ g mol}^{-1}} = 2\text{ mol} \)

Using formula (ii):
\( \text{Volume of the gas at STP} = \text{Number of moles of a gas (n)} \times \text{Molar volume of a gas} \)
\( = 2\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 44.8\text{ dm}^3 \)
This calculation assumes ideal gas behavior under standard temperature and pressure conditions.
Ans: Volume of ethane = \( 44.8\text{ dm}^3 \)
In simple words: First, we find how many moles are in 60 grams of ethane by dividing it by its molar mass (30 g/mol), which gives us 2 moles. Since 1 mole of any gas occupies 22.4 dm³ at STP, 2 moles will occupy twice that amount, which is 44.8 dm³.

🎯 Exam Tip: Always write down the given values, formulas, and units clearly to secure step-by-step marks in numerical problems.

Activity

Activity 1. Collect information of various scientists and prepare charts of their contribution in chemistry.
Answer:
Here is a summary of key scientists and their contributions:
1. John Dalton: Proposed the Atomic Theory, stating that all matter is made of indivisible atoms.
2. Amedeo Avogadro: Proposed Avogadro's Law, which states that equal volumes of gases at the same temperature and pressure contain an equal number of molecules.
3. Antoine Lavoisier: Known as the Father of Modern Chemistry, he established the Law of Conservation of Mass. Creating visual charts of these scientists helps us appreciate how modern chemical concepts were developed over time.
In simple words: This activity asks you to make a chart showing famous scientists like Dalton and Avogadro and what they discovered in chemistry.

🎯 Exam Tip: When preparing charts for activities, focus on the scientist's name, their main law/discovery, and the year of discovery to keep it concise and informative.

Question. ... chemistry.
Answer:

🎯 Exam Tip: When writing about scientists' contributions, make sure to list their specific discoveries, such as Gay-Lussac's gas law and Avogadro's work on atomic masses, to secure full marks.