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  4. Some Basic Concepts of Chemistry Chapter-Wise Test 11

Some Basic Concepts Of Chemistry Chapter-Wise Test 11

Correct answer Carries: 4.

Wrong Answer Carries: -1.

How many grams of \( \ce{Ca(OH)2} \) are required to produce 11.1 g of \( \ce{CaCl2} \) with excess \( \ce{HCl} \)? (Molar masses: \( \ce{Ca(OH)2} \) = 74 g/mol, \( \ce{CaCl2} \) = 111 g/mol)

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

Moles of \( \ce{CaCl2} \) = \( \frac{11.1}{111} \) = 0.1 mol.

1 mol \( \ce{CaCl2} \) from 1 mol \( \ce{Ca(OH)2} \); mass = 0.1 × 74 = 7.4 g.

3.7 g
14.8 g
7.4 g
11.1 g
3

What volume of \( \ce{CO2} \) at STP is produced when 15 g of \( \ce{C3H6} \) is burned completely? (Molar mass: \( \ce{C3H6} \) = 42 g/mol)

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

Moles of \( \ce{C3H6} \) = \( \frac{15}{42} \) ≈ 0.3571 mol.

2 mol \( \ce{C3H6} \) produce 6 mol \( \ce{CO2} \); 0.3571 mol produce \( \frac{6}{2} \) × 0.3571 ≈ 1.0714 mol.

Volume = 1.0714 × 22.4 ≈ 24 L.

12 L
8 L
16 L
24 L
4

How many significant figures are present in the result of \( \frac{0.0567 \times 273.15}{0.0821} \)?

0.0567 (3 sig figs), 273.15 (5 sig figs), 0.0821 (3 sig figs).

Result ≈ 188.56; limited to 3 sig figs (189).

2
4
3
5
3

A 0.78 g sample of a hydrocarbon produces 2.64 g of \( \ce{CO2} \) and 0.54 g of \( \ce{H2O} \) on complete combustion. What is its molecular formula if its molar mass is 78 g/mol? (Atomic masses: C = 12, H = 1, O = 16)

Mass of C = \( \frac{12}{44} \) × 2.64 ≈ 0.72 g; mass of H = \( \frac{2}{18} \) × 0.54 = 0.06 g.

Total = 0.72 + 0.06 = 0.78 g (matches).

Moles: C = \( \frac{0.72}{12} \) = 0.06, H = \( \frac{0.06}{1} \) = 0.06; ratio = 1 : 1; empirical formula = \( \ce{CH} \), mass = 13 g/mol.

\( n = \frac{78}{13} = 6 \); molecular formula = \( \ce{C6H6} \).

\( \ce{C6H6} \)
\( \ce{C5H10} \)
\( \ce{C4H8} \)
\( \ce{C3H6} \)
1

What volume of \( \ce{CO2} \) at STP is produced when 12 g of \( \ce{C2H4} \) is burned completely? (Molar mass: \( \ce{C2H4} \) = 28 g/mol)

Reaction: \( \ce{C2H4 + 3O2 -> 2CO2 + 2H2O} \).

Moles of \( \ce{C2H4} \) = \( \frac{12}{28} \) ≈ 0.4286 mol.

1 mol \( \ce{C2H4} \) produces 2 mol \( \ce{CO2} \); 0.4286 mol produces 0.8572 mol.

Volume = 0.8572 × 22.4 ≈ 19.2 L.

9.6 L
22.4 L
14.4 L
19.2 L
4

A solution of \( \ce{HCl} \) has a molarity of 2 M and a density of 1.06 g/mL. What is its molality? (Molar mass: \( \ce{HCl} \) = 36.5 g/mol)

Mass of 1 L solution = 1000 × 1.06 = 1060 g.

Mass of \( \ce{HCl} \) = 2 × 36.5 = 73 g.

Mass of water = 1060 - 73 = 987 g = 0.987 kg.

Molality = \( \frac{2}{0.987} \) ≈ 2.03 m.

2.03 m
1.89 m
2.15 m
1.75 m
1

What is the mass of \( \ce{PbCl2} \) produced when 27.8 g of \( \ce{Pb(NO3)2} \) reacts with excess \( \ce{NaCl} \)? (Molar masses: \( \ce{Pb(NO3)2} \) = 331 g/mol, \( \ce{PbCl2} \) = 278 g/mol)

Reaction: \( \ce{Pb(NO3)2 + 2NaCl -> PbCl2 + 2NaNO3} \).

Moles of \( \ce{Pb(NO3)2} \) = \( \frac{27.8}{331} \) ≈ 0.084 mol.

1 mol \( \ce{Pb(NO3)2} \) produces 1 mol \( \ce{PbCl2} \); mass = 0.084 × 278 ≈ 23.35 g.

13.9 g
33.1 g
18.5 g
23.35 g
4

How many grams of \( \ce{Al} \) are required to produce 5.6 L of \( \ce{H2} \) at STP with excess \( \ce{HCl} \)? (Atomic mass: Al = 27)

Reaction: \( \ce{2Al + 6HCl -> 2AlCl3 + 3H2} \).

Moles of \( \ce{H2} \) = \( \frac{5.6}{22.4} \) = 0.25 mol.

3 mol \( \ce{H2} \) from 2 mol Al; 0.25 mol from \( \frac{2}{3} \) × 0.25 ≈ 0.1667 mol Al.

Mass = 0.1667 × 27 ≈ 4.5 g.

2.7 g
4.5 g
9.0 g
13.5 g
2

What is the mass of \( \ce{PbSO4} \) produced when 20.7 g of \( \ce{Pb(NO3)2} \) reacts with excess \( \ce{H2SO4} \)? (Molar masses: \( \ce{Pb(NO3)2} \) = 331 g/mol, \( \ce{PbSO4} \) = 303 g/mol)

Reaction: \( \ce{Pb(NO3)2 + H2SO4 -> PbSO4 + 2HNO3} \).

Moles of \( \ce{Pb(NO3)2} \) = \( \frac{20.7}{331} \) ≈ 0.0625 mol.

1 mol \( \ce{Pb(NO3)2} \) produces 1 mol \( \ce{PbSO4} \); mass = 0.0625 × 303 ≈ 18.94 g.

15.15 g
9.47 g
24.3 g
18.94 g
4

A solution contains 15 ppm of \( \ce{CaCl2} \) by mass in water. What is its molarity if the density is 1 g/mL? (Molar mass: \( \ce{CaCl2} \) = 111 g/mol)

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

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

Moles = \( \frac{15}{111} \) ≈ 0.1351 mol; molarity = \( \frac{0.1351}{1000} \) ≈ 1.351 × 10⁻⁴ M.

1.2 × 10⁻⁴ M
1.5 × 10⁻⁴ M
1.351 × 10⁻⁴ M
1.0 × 10⁻⁴ M
3

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