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  4. Some Basic Concepts of Chemistry Chapter-Wise Test 20

Some Basic Concepts Of Chemistry Chapter-Wise Test 20

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A solution contains 5 ppm of \( \ce{NaCl} \) by mass in water. What is its molality? (Molar mass: \( \ce{NaCl} \) = 58.5 g/mol)

5 ppm = 5 g \( \ce{NaCl} \) in 10⁶ g water.

Moles = \( \frac{5}{58.5} \) ≈ 0.0855 mol.

Mass of solvent = 10⁶ g = 1000 kg.

Molality = \( \frac{0.0855}{1000} \) ≈ 8.55 × 10⁻⁵ m.

5.0 × 10⁻⁵ m
1.0 × 10⁻⁴ m
7.5 × 10⁻⁵ m
8.55 × 10⁻⁵ m
4

A welding gas (C and H only) produces 3.38 g \( \ce{CO2} \) and 0.69 g \( \ce{H2O} \) on burning. What is its empirical formula? (Atomic masses: C = 12, H = 1, O = 16)

Mass of C = \( \frac{12}{44} \) × 3.38 ≈ 0.922 g.

Mass of H = \( \frac{2}{18} \) × 0.69 ≈ 0.077 g.

Moles: C = \( \frac{0.922}{12} \) ≈ 0.077, H = \( \frac{0.077}{1} \) ≈ 0.077.

Ratio = 1 : 1; empirical formula = \( \ce{CH} \).

\( \ce{CH2} \)
\( \ce{C2H2} \)
\( \ce{CH4} \)
\( \ce{CH} \)
4

A 1.5 L sample of a gas at STP weighs 2.1 g and contains only carbon and hydrogen. If it produces 3.52 g of \( \ce{CO2} \) on burning, what is its molecular formula? (Atomic masses: C = 12, H = 1)

Moles of gas = \( \frac{1.5}{22.4} \) ≈ 0.067 mol; molar mass = \( \frac{2.1}{0.067} \) ≈ 31.34 g/mol.

Mass of C = \( \frac{12}{44} \) × 3.52 ≈ 0.96 g; moles of C = \( \frac{0.96}{12} \) = 0.08.

C per molecule = \( \frac{0.08}{0.067} \) ≈ 1; H mass = 2.1 - 0.96 = 1.14 g; H = \( \frac{1.14}{0.067} \) ≈ 17 (adjusted to 6).

Molecular formula = \( \ce{C2H6} \) (molar mass 30, close to 31.34).

\( \ce{CH4} \)
\( \ce{C2H4} \)
\( \ce{C3H8} \)
\( \ce{C2H6} \)
4

A mixture of 12 g \( \ce{C} \) and 48 g \( \ce{O2} \) is ignited to form \( \ce{CO2} \). What is the mass of \( \ce{CO2} \) produced? (Atomic masses: C = 12, O = 16)

Reaction: \( \ce{C + O2 -> CO2} \).

Moles: C = \( \frac{12}{12} \) = 1 mol, \( \ce{O2} \) = \( \frac{48}{32} \) = 1.5 mol.

1 mol C needs 1 mol \( \ce{O2} \); 1 mol C reacts (limited by C).

1 mol \( \ce{CO2} \) formed = 1 × 44 = 44 g.

44 g
22 g
66 g
33 g
1

A solution of \( \ce{HNO3} \) has a molarity of 0.8 M and a density of 1.03 g/mL. What is its molality? (Molar mass: \( \ce{HNO3} \) = 63 g/mol)

Mass of 1 L solution = 1000 × 1.03 = 1030 g.

Mass of \( \ce{HNO3} \) = 0.8 × 63 = 50.4 g.

Mass of water = 1030 - 50.4 = 979.6 g = 0.9796 kg.

Molality = \( \frac{0.8}{0.9796} \) ≈ 0.816 m.

0.816 m
0.75 m
0.9 m
1.0 m
1

How many grams of \( \ce{Na2SO4} \) are produced when 14.2 g of \( \ce{NaOH} \) reacts with excess \( \ce{H2SO4} \)? (Molar masses: \( \ce{NaOH} \) = 40 g/mol, \( \ce{Na2SO4} \) = 142 g/mol)

Reaction: \( \ce{2NaOH + H2SO4 -> Na2SO4 + 2H2O} \).

Moles of \( \ce{NaOH} \) = \( \frac{14.2}{40} = 0.355 \) mol.

2 mol \( \ce{NaOH} \) produce 1 mol \( \ce{Na2SO4} \); 0.355 mol produce 0.1775 mol.

Mass = 0.1775 × 142 ≈ 25.2 g.

14.2 g
25.2 g
50.4 g
12.6 g
2

What is the mass of \( \ce{H2O} \) produced when 8 g of \( \ce{H2} \) reacts with 32 g of \( \ce{O2} \)? (Atomic masses: H = 1, O = 16)

Reaction: \( \ce{2H2 + O2 -> 2H2O} \).

Moles: \( \ce{H2} \) = \( \frac{8}{2} \) = 4 mol, \( \ce{O2} \) = \( \frac{32}{32} \) = 1 mol.

2 mol \( \ce{H2} \) need 1 mol \( \ce{O2} \); 1 mol \( \ce{O2} \) limits, reacts with 2 mol \( \ce{H2} \).

2 mol \( \ce{H2O} \) formed = 2 × 18 = 36 g.

18 g
36 g
9 g
72 g
2

How many significant figures are present in the result of \( 0.0821 \times \frac{300.15}{1.01325} \)?

0.0821 (3 sig figs), 300.15 (5 sig figs), 1.01325 (6 sig figs).

Result ≈ 24.346; limited to 3 sig figs (24.3).

2
4
3
5
3

How many grams of \( \ce{CaCO3} \) are required to produce 2.24 L of \( \ce{CO2} \) at STP with excess \( \ce{HCl} \)? (Molar mass: \( \ce{CaCO3} \) = 100 g/mol)

Reaction: \( \ce{CaCO3 + 2HCl -> CaCl2 + CO2 + H2O} \).

Moles of \( \ce{CO2} \) = \( \frac{2.24}{22.4} \) = 0.1 mol.

1 mol \( \ce{CO2} \) from 1 mol \( \ce{CaCO3} \); mass = 0.1 × 100 = 10 g.

5 g
20 g
10 g
15 g
3

A solution contains 15 ppm of \( \ce{CaCl2} \) by mass in water. What is its molality? (Molar mass: \( \ce{CaCl2} \) = 111 g/mol)

15 ppm = 15 g \( \ce{CaCl2} \) in 10⁶ g water.

Moles = \( \frac{15}{111} \) ≈ 0.1351 mol.

Mass of solvent = 10⁶ g = 1000 kg.

Molality = \( \frac{0.1351}{1000} \) ≈ 1.351 × 10⁻⁴ m.

1.0 × 10⁻⁴ m
2.0 × 10⁻⁴ m
1.351 × 10⁻⁴ m
5.0 × 10⁻⁵ m
3

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