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  4. Some Basic Concepts of Chemistry Chapter-Wise Test 8

Some Basic Concepts Of Chemistry Chapter-Wise Test 8

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A mixture of 10 g \( \ce{CO} \) and 16 g \( \ce{O2} \) is ignited to form \( \ce{CO2} \). What is the mass of \( \ce{CO2} \) produced? (Atomic masses: C = 12, O = 16)

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

Moles: \( \ce{CO} \) = \( \frac{10}{28} \) ≈ 0.357 mol, \( \ce{O2} \) = \( \frac{16}{32} \) = 0.5 mol.

2 mol \( \ce{CO} \) need 1 mol \( \ce{O2} \); 0.357 mol need 0.1785 mol \( \ce{O2} \) (available).

0.357 mol \( \ce{CO} \) produces 0.357 mol \( \ce{CO2} \) = 0.357 × 44 ≈ 15.71 g.

15.71 g
22 g
11 g
44 g
1

How many grams of \( \ce{Zn} \) are required to produce 4.48 L of \( \ce{H2} \) at STP with excess \( \ce{HCl} \)? (Atomic mass: Zn = 65)

Reaction: \( \ce{Zn + 2HCl -> ZnCl2 + H2} \).

Moles of \( \ce{H2} \) = \( \frac{4.48}{22.4} \) = 0.2 mol.

1 mol \( \ce{H2} \) from 1 mol Zn; mass = 0.2 × 65 = 13 g.

6.5 g
19.5 g
9.75 g
13 g
4

A solution contains 25 ppm of \( \ce{NaNO3} \) by mass in water. What is its molarity if the density is 1 g/mL? (Molar mass: \( \ce{NaNO3} \) = 85 g/mol)

25 ppm = 25 g \( \ce{NaNO3} \) in 10⁶ g solution.

Volume = \( \frac{10⁶}{1} \) = 10⁶ mL = 1000 L.

Moles = \( \frac{25}{85} \) ≈ 0.2941 mol; molarity = \( \frac{0.2941}{1000} \) ≈ 2.941 × 10⁻⁴ M.

2.0 × 10⁻⁴ M
3.5 × 10⁻⁴ M
2.941 × 10⁻⁴ M
1.5 × 10⁻⁴ M
3

What mass of \( \ce{Fe} \) can be obtained from 16 g of \( \ce{Fe2O3} \) using excess \( \ce{CO} \)? (Molar masses: \( \ce{Fe2O3} \) = 160 g/mol, Fe = 56 g/mol)

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

Moles of \( \ce{Fe2O3} \) = \( \frac{16}{160} \) = 0.1 mol.

0.1 mol \( \ce{Fe2O3} \) produces 0.2 mol Fe = 0.2 × 56 = 11.2 g.

5.6 g
22.4 g
11.2 g
8.4 g
3

What is the mass percentage of \( \ce{NH3} \) in a solution made by dissolving 3.4 g of \( \ce{NH3} \) in 16.6 g of water? (Molar mass: \( \ce{NH3} \) = 17 g/mol)

Total mass = 3.4 + 16.6 = 20 g.

Mass % = \( \frac{3.4}{20} \) × 100 = 17%.

15%
20%
17%
25%
3

A solution of \( \ce{KOH} \) has a density of 1.1 g/mL and a molarity of 2 M. What is its molality? (Molar mass: \( \ce{KOH} \) = 56 g/mol)

Mass of 1 L solution = 1000 × 1.1 = 1100 g.

Mass of \( \ce{KOH} \) = 2 × 56 = 112 g.

Mass of water = 1100 - 112 = 988 g = 0.988 kg.

Molality = \( \frac{2}{0.988} \) ≈ 2.02 m.

2.02 m
1.82 m
2.20 m
1.50 m
1

A mixture of 8 g \( \ce{CH4} \) and 32 g \( \ce{O2} \) is ignited to form \( \ce{CO2} \) and \( \ce{H2O} \). What is the mass of \( \ce{CO2} \) produced? (Atomic masses: C = 12, H = 1, O = 16)

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

Moles: \( \ce{CH4} \) = \( \frac{8}{16} \) = 0.5 mol, \( \ce{O2} \) = \( \frac{32}{32} \) = 1 mol.

1 mol \( \ce{CH4} \) needs 2 mol \( \ce{O2} \); 0.5 mol needs 1 mol \( \ce{O2} \) (available).

0.5 mol \( \ce{CO2} \) formed = 0.5 × 44 = 22 g.

22 g
44 g
11 g
33 g
1

How many grams of \( \ce{ZnSO4} \) are required to produce 6.5 g of \( \ce{Zn} \) with excess \( \ce{H2SO4} \)? (Molar masses: \( \ce{ZnSO4} \) = 161.5 g/mol, Zn = 65 g/mol)

Reaction: \( \ce{ZnSO4 + H2SO4 -> Zn + H2 + SO3} \) (simplified context).

Moles of Zn = \( \frac{6.5}{65} \) = 0.1 mol.

1 mol Zn from 1 mol \( \ce{ZnSO4} \); mass = 0.1 × 161.5 = 16.15 g.

8.075 g
16.15 g
32.3 g
12.1 g
2

A 0.1 M \( \ce{NaOH} \) solution is prepared using 4 g of \( \ce{NaOH} \). What is the volume of the solution in liters? (Molar mass: \( \ce{NaOH} \) = 40 g/mol)

Moles of \( \ce{NaOH} \) = \( \frac{4}{40} \) = 0.1 mol.

Molarity = \( \frac{\text{moles}}{\text{volume}} \); 0.1 = \( \frac{0.1}{V} \); \( V = 1 \) L.

0.5 L
2 L
1 L
0.25 L
3

A gas occupies 4.48 L at STP and weighs 2 g. What is its molar mass?

Moles = \( \frac{4.48}{22.4} \) = 0.2 mol.

Molar mass = \( \frac{2}{0.2} \) = 10 g/mol.

20 g/mol
5 g/mol
10 g/mol
40 g/mol
3

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