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  4. Some Basic Concepts of Chemistry Chapter-Wise Test 2

Some Basic Concepts Of Chemistry Chapter-Wise Test 2

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A solution of \( \ce{H2SO4} \) has a molarity of 0.5 M and a density of 1.03 g/mL. What is its molality? (Molar mass: \( \ce{H2SO4} \) = 98 g/mol)

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

Mass of \( \ce{H2SO4} \) = 0.5 × 98 = 49 g.

Mass of water = 1030 - 49 = 981 g = 0.981 kg.

Molality = \( \frac{0.5}{0.981} \) ≈ 0.51 m.

0.51 m
0.47 m
0.55 m
0.60 m
1

A 10 L sample of a gas at STP weighs 11.6 g and contains only C and H. If it produces 3.38 g \( \ce{CO2} \) on burning, what is its molecular formula? (Atomic masses: C = 12, H = 1)

Moles of gas = \( \frac{10}{22.4} \) ≈ 0.446 mol; molar mass = \( \frac{11.6}{0.446} \) ≈ 26 g/mol.

Mass of C = \( \frac{12}{44} \) × 3.38 ≈ 0.922 g; moles of C = \( \frac{0.922}{12} \) ≈ 0.077.

0.446 mol gas has 0.077 mol C; C atoms per molecule = \( \frac{0.077}{0.446} \) ≈ 0.172 × 6 ≈ 1.

H mass = 11.6 - 0.922 × 6 = 6.07 g; H atoms = \( \frac{6.07}{1} \) ÷ 0.446 ≈ 13.6 ≈ 14.

Molecular formula = \( \ce{CH14} \) (adjusted to \( \ce{C2H2} \), molar mass 26).

\( \ce{CH4} \)
\( \ce{C2H4} \)
\( \ce{CH2} \)
\( \ce{C2H2} \)
4

How many grams of \( \ce{CuSO4} \) are required to produce 12.8 g of \( \ce{Cu} \) with excess \( \ce{Zn} \)? (Molar masses: \( \ce{CuSO4} \) = 159.5 g/mol, Cu = 64 g/mol)

Reaction: \( \ce{CuSO4 + Zn -> Cu + ZnSO4} \).

Moles of Cu = \( \frac{12.8}{64} \) = 0.2 mol.

1 mol Cu from 1 mol \( \ce{CuSO4} \); mass = 0.2 × 159.5 = 31.9 g.

15.95 g
31.9 g
63.8 g
47.85 g
2

A 250 mL solution contains 9.75 g of \( \ce{HNO3} \) and has a density of 1.03 g/mL. What is its molarity? (Molar mass: \( \ce{HNO3} \) = 63 g/mol)

Moles of \( \ce{HNO3} \) = \( \frac{9.75}{63} \) ≈ 0.1548 mol.

Volume = 250 mL = 0.25 L.

Molarity = \( \frac{0.1548}{0.25} \) ≈ 0.619 M.

0.5 M
0.619 M
1.0 M
0.75 M
2

How many significant figures are present in the result of \( \frac{0.0256 \times 273.15}{0.0821} \)?

0.0256 (3 sig figs), 273.15 (5 sig figs), 0.0821 (3 sig figs).

Result ≈ 85.16; limited to 3 sig figs (85.2).

2
4
3
5
3

How many grams of \( \ce{MgCl2} \) are produced when 9.5 g of \( \ce{Mg(OH)2} \) reacts with excess \( \ce{HCl} \)? (Molar masses: \( \ce{Mg(OH)2} \) = 58 g/mol, \( \ce{MgCl2} \) = 95 g/mol)

Reaction: \( \ce{Mg(OH)2 + 2HCl -> MgCl2 + 2H2O} \).

Moles of \( \ce{Mg(OH)2} \) = \( \frac{9.5}{58} \) ≈ 0.1638 mol.

1 mol \( \ce{Mg(OH)2} \) produces 1 mol \( \ce{MgCl2} \); mass = 0.1638 × 95 ≈ 15.56 g.

7.78 g
15.56 g
31.12 g
12.35 g
1

A 0.15 M \( \ce{KOH} \) solution is diluted by adding 150 mL of water to 50 mL of the original solution. What is the final molarity?

\( M_1 V_1 = M_2 V_2 \).

0.15 × 50 = \( M_2 \) × (50 + 150).

\( M_2 = \frac{0.15 \times 50}{200} = 0.0375 \) M.

0.05 M
0.1 M
0.075 M
0.0375 M
4

What is the mass percentage of \( \ce{KOH} \) in a solution made by dissolving 5.6 g of \( \ce{KOH} \) in 44.4 g of water? (Molar mass: \( \ce{KOH} \) = 56 g/mol)

Total mass = 5.6 + 44.4 = 50 g.

Mass % = \( \frac{5.6}{50} \) × 100 = 11.2%.

10%
15%
11.2%
20%
3

What is the mole fraction of \( \ce{NaCl} \) in a solution made by dissolving 5.85 g of \( \ce{NaCl} \) in 90 g of water? (Molar masses: \( \ce{NaCl} \) = 58.5 g/mol, \( \ce{H2O} \) = 18 g/mol)

Moles of \( \ce{NaCl} \) = \( \frac{5.85}{58.5} \) = 0.1 mol.

Moles of \( \ce{H2O} \) = \( \frac{90}{18} \) = 5 mol.

Total moles = 0.1 + 5 = 5.1; mole fraction = \( \frac{0.1}{5.1} \) ≈ 0.0196.

0.1
0.0196
0.05
0.0098
2

What volume of \( \ce{O2} \) at STP is required to burn 7.8 g of \( \ce{C3H6} \) completely to \( \ce{CO2} \) and \( \ce{H2O} \)? (Molar mass: \( \ce{C3H6} \) = 42 g/mol)

Reaction: \( \ce{2C3H6 + 9O2 -> 6CO2 + 6H2O} \).

Moles of \( \ce{C3H6} \) = \( \frac{7.8}{42} \) ≈ 0.1857 mol.

2 mol \( \ce{C3H6} \) need 9 mol \( \ce{O2} \); 0.1857 mol need \( \frac{9}{2} \) × 0.1857 ≈ 0.8357 mol.

Volume = 0.8357 × 22.4 ≈ 18.72 L.

18.72 L
9.36 L
22.4 L
12.48 L
1

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