Chapter 1: Some Basic Concepts of Chemistry
Table of Contents
1.3. Properties of Matter and their Measurement
1. Weight (g) of two moles of the organic compound, which is obtained by heating sodium ethanoate with sodium hydroxide in presence of calcium oxide is:
2. 0.24 g of a volatile gas, upon vapourisation, gives 45 mL vapour at NTP. What will be the vapour density of the substance? (Density of \( H_{2} = 0.089 \))
3. The dimensions of pressure are the same as that of:
1.4. Uncertainty in Measurement
4. Given the numbers: 161 cm, 0.161 cm, 0.0161 cm. The number of significant figures for the three numbers are:
5. In the final answer of the expression:
the number of significant figures is:
1.5. Laws of Chemical Combinations
6. Equal masses of \( H_{2} \), \( O_{2} \) and methane have been taken in a container of volume V at temperature 27°C in identical condition. The ratio of the volumes of gases \( H_{2} : O_{2} : CH_{4} \) would be:
7. The molecular weight of \( O_{2} \) and \( SO_{2} \) are 32 g and 64 g respectively. At \( 15^{\circ}C \) and 150 mmHg pressure, one litre of \( O_{2} \) contains \( N \) molecules. The number of molecules in two litres of \( SO_{2} \) under the same conditions of temperature and pressure will be:
1.7. Atomic and Molecular Masses
8. Suppose the elements X and Y combine to form two compounds \( XY_{2} \) and \( X_{3}Y_{2} \). When 0.1 mole of \( XY_{2} \) weighs 10 g and 0.05 mole of \( X_{3}Y_{2} \) weighs 9 g, the atomic weights of X and Y are:
9. An element X has the following isotopic composition: \( {}^{200}X: 90\% \), \( {}^{199}X: 8.0\% \), \( {}^{202}X: 2.0\% \). The weighted average atomic mass of the naturally-occurring element X is closest to:
10. Boron has two stable isotopes, \( {}^{10}B \) (19%) and \( {}^{11}B \) (81%). Calculate average atomic weight of boron in the periodic table.
1.8. Mole Concept and Molar Masses
12. 1.0 g of \( H_{2} \) has same number of molecules as in:
21. A mixture of gases contains \( H_{2} \) and \( O_{2} \) gases in the ratio of 1:4 (w/w). What is the molar ratio of the two gases in the mixture?
24. Volume occupied by one molecule of water (density = 1 g \( cm^{-3} \)) is:
1.9. Percentage Composition
37. An organic compound contains 78% (by wt.) carbon and remaining percentage of hydrogen. The right option for the empirical formula of this compound is: (Atomic wt. of C is 12, H is 1)
38. Magnesium reacts with an element (X) to form an ionic compound. If the ground state electronic configuration of (X) is \( 1s^{2} 2s^{2} 2p^{3} \), the simplest formula for this compound is:
1.10. Stoichiometry and Stoichiometric Calculations
45. What mass of 95% pure \( CaCO_{3} \) will be required to neutralize 50 mL of 0.5 M HCl solution according to the following reaction?
64. A 5 molar solution of \( H_{2}SO_{4} \) is diluted from 1 litre to a volume of 10 litres, the normality of the solution will be:
Solutions and Explanations
Solution 1
(D) Heating sodium ethanoate with soda lime (NaOH + CaO) produces methane (\( CH_{4} \)).
One mole of \( CH_{4} = 16 \, \text{g} \). Two moles of \( CH_{4} = 32 \, \text{g} \).
Solution 2
(B) Given: \( \text{Density of } H_{2} = 0.089 \). Volume = 45 mL or 0.045 L.
\( \text{Weight of } H_{2} = \text{Volume} \times \text{Density} = 0.045 \times 0.089 = 0.004005 \, \text{g} \).
\( \text{Vapour Density} = \frac{\text{Weight of substance}}{\text{Weight of same volume of } H_{2}} = \frac{0.24}{0.004005} \approx 59.93 \).
Solution 4
| Numbers | Significant Figures |
|---|---|
| 161 cm | 3 |
| 0.161 cm | 3 |
| 0.0161 cm | 3 |
Solution 8
(A) Let atomic masses be \( A_{X} \) and \( A_{Y} \).
- For \( XY_{2} \): \( 0.1 = \frac{10}{A_{X} + 2A_{Y}} \Rightarrow A_{X} + 2A_{Y} = 100 \)
- For \( X_{3}Y_{2} \): \( 0.05 = \frac{9}{3A_{X} + 2A_{Y}} \Rightarrow 3A_{X} + 2A_{Y} = 180 \)
Solving these gives \( A_{X} = 40 \, \text{g/mol} \) and \( A_{Y} = 30 \, \text{g/mol} \).
Solution 39
Empirical formula calculation for C, H, O:
| Element | Percentage | Molar Ratio | Simplest Ratio |
|---|---|---|---|
| C | 38.71 | 3.22 | 1 |
| H | 9.67 | 9.67 | 3 |
| O | 51.62 | 3.22 | 1 |
Empirical Formula: \( CH_{3}O \).
Solution 64
(A) For dilution: \( N_{1}V_{1} = N_{2}V_{2} \).
\( \text{Normality} = \text{Molarity} \times \text{Basicity} \). For \( H_{2}SO_{4} \), basicity = 2.
\( N_{1} = 5 \, \text{M} \times 2 = 10 \, \text{N} \). \( V_{1} = 1 \, \text{L} \), \( V_{2} = 10 \, \text{L} \).
\( 10 \times 1 = N_{2} \times 10 \Rightarrow N_{2} = 1 \, \text{N} \).
Chapter 1: Some Basic Concepts of Chemistry
Table of Contents
1.3. Properties of Matter and their Measurement
1.4. Uncertainty in Measurement
1.5. Laws of Chemical Combinations
1.7. Atomic and Molecular Masses
1.8. Mole Concept and Molar Masses
1.9. Percentage Composition
1.10. Stoichiometry and Stoichiometric Calculations
Solutions and Explanations
Solution 1
Answer: (D)
Heating sodium ethanoate with soda lime (NaOH + CaO) produces methane (\( CH_{4} \)).
One mole of \( CH_{4} \) weighs 16 g. Therefore, two moles of \( CH_{4} \) will weigh 32 g.
Solution 2
Answer: (B)
Given: \( H_{2} \) density = 0.089. Volume = 45 mL or 0.045 L.
Weight of \( H_{2} = \text{Volume} \times \text{Density} = 0.045 \times 0.089 = 0.004005 \) g.
Vapour Density = \( \frac{\text{Weight of substance}}{\text{Weight of same volume of hydrogen}} = \frac{0.24}{0.004005} \approx 59.93 \).
Solution 4
Answer: (D)
| Numbers | Significant Figures |
|---|---|
| 161 cm | 3 |
| 0.161 cm | 3 |
| 0.0161 cm | 3 |
Solution 8
Answer: (A)
Let atomic masses of X and Y be \( A_{X} \) and \( A_{Y} \).
For \( XY_{2} \): \( 0.1 = \frac{10}{A_{X} + 2A_{Y}} \implies A_{X} + 2A_{Y} = 100 \)
For \( X_{3}Y_{2} \): \( 0.05 = \frac{9}{3A_{X} + 2A_{Y}} \implies 3A_{X} + 2A_{Y} = 180 \)
Solving the equations yields \( A_{X} = 40 \) and \( A_{Y} = 30 \).
Solution 39
| Element | Percentage | Molar Ratio | Simplest Molar Ratio |
|---|---|---|---|
| C | 38.71 | 3.22 | 1 |
| H | 9.67 | 9.67 | 3 |
| O | 51.62 | 3.22 | 1 |
Empirical formula: \( CH_{3}O \).
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