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  4. Some Basic Concepts of Chemistry Chapter-Wise Test 13

Some Basic Concepts Of Chemistry Chapter-Wise Test 13

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A 0.5 M \( \ce{HNO3} \) solution is diluted from 100 mL to 400 mL. What is the final molarity?

\( M_1 V_1 = M_2 V_2 \).

0.5 × 100 = \( M_2 \) × 400.

\( M_2 = \frac{0.5 \times 100}{400} = 0.125 \) M.

0.2 M
0.25 M
0.125 M
0.5 M
3

How many significant figures are present in the result of \( \frac{0.0456 \times 298.15}{0.112} \)?

0.0456 (3 sig figs), 298.15 (5 sig figs), 0.112 (3 sig figs).

Result = \( \frac{0.0456 \times 298.15}{0.112} \) ≈ 121.36; limited to 3 sig figs (121).

2
4
3
5
3

A mixture of 16 g \( \ce{CH4} \) and 32 g \( \ce{O2} \) is ignited. What is the mass of \( \ce{CO2} \) produced? (Atomic masses: C = 12, H = 1, O = 16)

Reaction: \( \ce{CH4 + 2O2 -> CO2 + 2H2O} \).

Moles: \( \ce{CH4} \) = \( \frac{16}{16} \) = 1 mol, \( \ce{O2} \) = \( \frac{32}{32} \) = 1 mol.

1 mol \( \ce{CH4} \) needs 2 mol \( \ce{O2} \); \( \ce{O2} \) is limiting (1 mol).

0.5 mol \( \ce{CH4} \) reacts, producing 0.5 mol \( \ce{CO2} \) = 22 g.

44 g
22 g
11 g
33 g
2

A solution of \( \ce{KOH} \) has a molarity of 1.5 M and a density of 1.05 g/mL. What is its molality? (Molar mass: \( \ce{KOH} \) = 56 g/mol)

Mass of 1 L solution = 1000 × 1.05 = 1050 g.

Mass of \( \ce{KOH} \) = 1.5 × 56 = 84 g.

Mass of water = 1050 - 84 = 966 g = 0.966 kg.

Molality = \( \frac{1.5}{0.966} \) ≈ 1.55 m.

1.55 m
1.45 m
1.65 m
1.75 m
1

How many grams of \( \ce{Fe2O3} \) are required to produce 11.2 g of \( \ce{Fe} \) with excess \( \ce{CO} \)? (Molar masses: \( \ce{Fe2O3} \) = 160 g/mol, Fe = 56 g/mol)

Reaction: \( \ce{Fe2O3 + 3CO -> 2Fe + 3CO2} \).

Moles of Fe = \( \frac{11.2}{56} \) = 0.2 mol.

2 mol Fe from 1 mol \( \ce{Fe2O3} \); 0.2 mol from 0.1 mol; mass = 0.1 × 160 = 16 g.

8 g
32 g
24 g
16 g
4

What is the mass of \( \ce{AgNO3} \) produced when 14.3 g of \( \ce{AgCl} \) reacts with excess \( \ce{HNO3} \)? (Molar masses: \( \ce{AgCl} \) = 143.5 g/mol, \( \ce{AgNO3} \) = 170 g/mol)

Reaction: \( \ce{AgCl + HNO3 -> AgNO3 + HCl} \).

Moles of \( \ce{AgCl} \) = \( \frac{14.3}{143.5} \) ≈ 0.0997 mol.

1 mol \( \ce{AgCl} \) produces 1 mol \( \ce{AgNO3} \); mass = 0.0997 × 170 ≈ 16.95 g.

8.48 g
25.44 g
12.71 g
16.95 g
4

How many grams of \( \ce{MnO2} \) are required to produce 1.42 g of \( \ce{Cl2} \) with excess \( \ce{HCl} \)? (Molar masses: \( \ce{MnO2} \) = 87 g/mol, \( \ce{Cl2} \) = 71 g/mol)

Reaction: \( \ce{MnO2 + 4HCl -> MnCl2 + 2H2O + Cl2} \).

Moles of \( \ce{Cl2} \) = \( \frac{1.42}{71} \) = 0.02 mol.

1 mol \( \ce{Cl2} \) needs 1 mol \( \ce{MnO2} \); 0.02 mol needs 0.02 × 87 = 1.74 g.

0.87 g
3.48 g
2.61 g
1.74 g
4

How many grams of \( \ce{NaOH} \) are required to produce 14.6 g of \( \ce{NaNO3} \) with excess \( \ce{HNO3} \)? (Molar masses: \( \ce{NaOH} \) = 40 g/mol, \( \ce{NaNO3} \) = 85 g/mol)

Reaction: \( \ce{NaOH + HNO3 -> NaNO3 + H2O} \).

Moles of \( \ce{NaNO3} \) = \( \frac{14.6}{85} \) ≈ 0.1718 mol.

1 mol \( \ce{NaNO3} \) from 1 mol \( \ce{NaOH} \); mass = 0.1718 × 40 ≈ 6.87 g.

3.43 g
10.3 g
6.87 g
8.5 g
3

How many grams of \( \ce{MgCO3} \) are required to produce 4.4 g of \( \ce{CO2} \) with excess \( \ce{HCl} \)? (Molar masses: \( \ce{MgCO3} \) = 84 g/mol, \( \ce{CO2} \) = 44 g/mol)

Reaction: \( \ce{MgCO3 + 2HCl -> MgCl2 + CO2 + H2O} \).

Moles of \( \ce{CO2} \) = \( \frac{4.4}{44} \) = 0.1 mol.

1 mol \( \ce{CO2} \) from 1 mol \( \ce{MgCO3} \); mass = 0.1 × 84 = 8.4 g.

4.2 g
8.4 g
12.6 g
16.8 g
2

A mixture of 12 g \( \ce{C} \) and 40 g \( \ce{O2} \) is ignited to form \( \ce{CO2} \) and \( \ce{CO} \). If 22 g of \( \ce{CO2} \) is produced, what is the mass of \( \ce{CO} \) formed? (Atomic masses: C = 12, O = 16)

Moles: C = \( \frac{12}{12} \) = 1 mol, \( \ce{O2} \) = \( \frac{40}{32} \) = 1.25 mol.

Moles of \( \ce{CO2} \) = \( \frac{22}{44} \) = 0.5 mol; C used = 0.5 mol, O used = 0.5 mol.

Remaining: C = 1 - 0.5 = 0.5 mol, O = 1.25 - 0.5 = 0.75 mol (as atoms, 1.5 mol).

0.5 mol C forms 0.5 mol \( \ce{CO} \) with 0.5 mol O; mass = 0.5 × 28 = 14 g.

14 g
28 g
7 g
42 g
1

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