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  4. Some Basic Concepts of Chemistry Chapter-Wise Test 17

Some Basic Concepts Of Chemistry Chapter-Wise Test 17

Correct answer Carries: 4.

Wrong Answer Carries: -1.

How many grams of \( \ce{Na2CO3} \) are required to produce 8.4 g of \( \ce{NaCl} \) with excess \( \ce{HCl} \)? (Molar masses: \( \ce{Na2CO3} \) = 106 g/mol, \( \ce{NaCl} \) = 58.5 g/mol)

Reaction: \( \ce{Na2CO3 + 2HCl -> 2NaCl + CO2 + H2O} \).

Moles of \( \ce{NaCl} \) = \( \frac{8.4}{58.5} \) ≈ 0.1436 mol.

2 mol \( \ce{NaCl} \) from 1 mol \( \ce{Na2CO3} \); 0.1436 mol from 0.0718 mol.

Mass = 0.0718 × 106 ≈ 7.61 g.

5.3 g
10.6 g
7.61 g
15.2 g
3

A mixture of 6 g \( \ce{H2} \) and 44 g \( \ce{O2} \) is ignited to form \( \ce{H2O} \). What is the mass of \( \ce{H2O} \) produced? (Atomic masses: H = 1, O = 16)

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

Moles: \( \ce{H2} \) = \( \frac{6}{2} \) = 3 mol, \( \ce{O2} \) = \( \frac{44}{32} \) = 1.375 mol.

2 mol \( \ce{H2} \) need 1 mol \( \ce{O2} \); 1.375 mol \( \ce{O2} \) limits, reacts with 2.75 mol \( \ce{H2} \).

2.75 mol \( \ce{H2O} \) formed = 2.75 × 18 = 49.5 g.

49.5 g
36 g
54 g
18 g
1

What volume of \( \ce{CO2} \) at STP is produced when 10 g of \( \ce{C2H2} \) is burned completely? (Molar mass: \( \ce{C2H2} \) = 26 g/mol)

Reaction: \( \ce{2C2H2 + 5O2 -> 4CO2 + 2H2O} \).

Moles of \( \ce{C2H2} \) = \( \frac{10}{26} \) ≈ 0.3846 mol.

2 mol \( \ce{C2H2} \) produce 4 mol \( \ce{CO2} \); 0.3846 mol produce 0.7692 mol.

Volume = 0.7692 × 22.4 ≈ 17.23 L.

8.61 L
22.4 L
11.2 L
17.23 L
4

A 3 M solution of \( \ce{NaCl} \) has a density of 1.25 g/mL. What is its molality? (Molar mass of \( \ce{NaCl} \) = 58.5 g/mol)

Mass of 1 L solution = 1000 × 1.25 = 1250 g.

Mass of \( \ce{NaCl} \) = 3 × 58.5 = 175.5 g.

Mass of water = 1250 - 175.5 = 1074.5 g = 1.0745 kg.

Molality = \( \frac{3}{1.0745} \) ≈ 2.79 m.

2.79 m
3.00 m
2.50 m
3.25 m
1

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 of solution = \( 3.4 + 16.6 = 20.0 \) g (3 significant figures).

Mass percentage of \( \ce{NH3} \) = \( \frac{\text{mass of } \ce{NH3}}{\text{total mass}} \times 100 = \frac{3.4}{20.0} \times 100 = 17.0\% \) (3 significant figures).

The molar mass of \( \ce{NH3} \) is not needed for mass percentage calculation.

A. 15%
B. 20%
C. 17%
D. 25%
3

A solution of \( \ce{H2SO4} \) has a molarity of 0.5 M and a density of 1.02 g/mL. What is its molality? (Molar mass: \( \ce{H2SO4} \) = 98 g/mol)

Mass of 1 L solution = 1000 × 1.02 = 1020 g.

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

Mass of water = 1020 - 49 = 971 g = 0.971 kg.

Molality = \( \frac{0.5}{0.971} \) ≈ 0.515 m.

0.515 m
0.49 m
0.55 m
0.6 m
1

How many grams of \( \ce{KMnO4} \) are required to oxidize 9.8 g of \( \ce{FeSO4} \) in acidic medium to \( \ce{Fe2(SO4)3} \)? (Molar masses: \( \ce{KMnO4} \) = 158 g/mol, \( \ce{FeSO4} \) = 152 g/mol)

Reaction: \( \ce{2KMnO4 + 10FeSO4 + 8H2SO4 -> 5Fe2(SO4)3 + 2MnSO4 + K2SO4 + 8H2O} \).

Moles of \( \ce{FeSO4} \) = \( \frac{9.8}{152} \) ≈ 0.0645 mol.

10 mol \( \ce{FeSO4} \) need 2 mol \( \ce{KMnO4} \); 0.0645 mol needs \( \frac{2}{10} \) × 0.0645 = 0.0129 mol.

Mass = 0.0129 × 158 ≈ 2.04 g.

1.58 g
2.04 g
3.16 g
4.08 g
1

How many grams of \( \ce{Zn} \) are required to produce 2.24 L of \( \ce{H2} \) at STP with excess \( \ce{H2SO4} \)? (Atomic mass: Zn = 65)

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

Moles of \( \ce{H2} \) = \( \frac{2.24}{22.4} \) = 0.1 mol.

1 mol \( \ce{H2} \) from 1 mol Zn; mass = 0.1 × 65 = 6.5 g.

3.25 g
13 g
9.75 g
6.5 g
4

What is the mass of \( \ce{AgCl} \) produced when 17 g of \( \ce{AgNO3} \) reacts with excess \( \ce{NaCl} \)? (Molar masses: \( \ce{AgNO3} \) = 170 g/mol, \( \ce{AgCl} \) = 143.5 g/mol)

Reaction: \( \ce{AgNO3 + NaCl -> AgCl + NaNO3} \).

Moles of \( \ce{AgNO3} \) = \( \frac{17}{170} \) = 0.1 mol.

1 mol \( \ce{AgNO3} \) produces 1 mol \( \ce{AgCl} \); mass = 0.1 × 143.5 = 14.35 g.

7.15 g
28.7 g
10.5 g
14.35 g
4

A 500 mL solution of \( \ce{H2SO4} \) has a molarity of 0.2 M and density 1.02 g/mL. What is the mass percentage of \( \ce{H2SO4} \)? (Molar mass: \( \ce{H2SO4} \) = 98 g/mol)

Moles = 0.2 × 0.5 = 0.1 mol; mass of \( \ce{H2SO4} \) = 0.1 × 98 = 9.8 g.

Mass of solution = 500 × 1.02 = 510 g.

Mass % = \( \frac{9.8}{510} \) × 100 ≈ 1.92%.

0.98%
1.92%
2.45%
3.84%
2

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