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  4. Some Basic Concepts of Chemistry Chapter-Wise Test 9

Some Basic Concepts Of Chemistry Chapter-Wise Test 9

Correct answer Carries: 4.

Wrong Answer Carries: -1.

How many moles of \( \ce{CH4} \) are required to produce 88 g of \( \ce{CO2} \) if the reaction yield is 50%? (Molar mass: \( \ce{CH4} \) = 16 g/mol, \( \ce{CO2} \) = 44 g/mol)

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

Moles of \( \ce{CO2} \) = \( \frac{88}{44} \) = 2 mol.

With 50% yield, actual \( \ce{CH4} \) moles = \( \frac{2}{0.5} \) = 4 mol.

4
2
1
8
1

A compound contains 26.67% carbon, 2.22% hydrogen, and 71.11% oxygen by mass. If its molar mass is 90 g/mol, what is its molecular formula? (Atomic masses: C = 12, H = 1, O = 16)

For 100 g: C = 26.67 g, H = 2.22 g, O = 71.11 g.

Moles: C = \( \frac{26.67}{12} \) ≈ 2.22, H = \( \frac{2.22}{1} \) = 2.22, O = \( \frac{71.11}{16} \) ≈ 4.44.

Ratio = 1 : 1 : 2; empirical formula = \( \ce{CHO2} \), mass = 45 g/mol.

\( n = \frac{90}{45} = 2 \); molecular formula = \( \ce{C2H2O4} \).

\( \ce{C2H2O4} \)
\( \ce{CH2O2} \)
\( \ce{C3H3O6} \)
\( \ce{CHO2} \)
1

A solution contains 20 ppm of \( \ce{MgCl2} \) by mass in water. What is its molality? (Molar mass: \( \ce{MgCl2} \) = 95 g/mol)

20 ppm = 20 g \( \ce{MgCl2} \) in 10⁶ g water.

Moles = \( \frac{20}{95} \) ≈ 0.2105 mol.

Mass of solvent = 10⁶ g = 1000 kg.

Molality = \( \frac{0.2105}{1000} \) ≈ 2.105 × 10⁻⁴ m.

2.0 × 10⁻⁴ m
2.5 × 10⁻⁴ m
2.105 × 10⁻⁴ m
1.5 × 10⁻⁴ m
3

A compound contains 40% carbon, 6.67% hydrogen, and 53.33% oxygen by mass. What is its empirical formula? (Atomic masses: C = 12, H = 1, O = 16)

For 100 g: C = 40 g, H = 6.67 g, O = 53.33 g.

Moles: C = \( \frac{40}{12} \) ≈ 3.33, H = \( \frac{6.67}{1} \) ≈ 6.67, O = \( \frac{53.33}{16} \) ≈ 3.33.

Ratio = 1 : 2 : 1; empirical formula = \( \ce{CH2O} \).

\( \ce{CHO} \)
\( \ce{CH2O} \)
\( \ce{C2H4O2} \)
\( \ce{CH3O} \)
2

What is the mass of \( \ce{BaSO4} \) produced when 20 g of \( \ce{BaCl2} \) reacts with excess \( \ce{H2SO4} \)? (Molar masses: \( \ce{BaCl2} \) = 208 g/mol, \( \ce{BaSO4} \) = 233 g/mol)

Reaction: \( \ce{BaCl2 + H2SO4 -> BaSO4 + 2HCl} \).

Moles of \( \ce{BaCl2} \) = \( \frac{20}{208} \) ≈ 0.0962 mol.

1 mol \( \ce{BaCl2} \) produces 1 mol \( \ce{BaSO4} \); mass = 0.0962 × 233 ≈ 22.41 g.

11.2 g
22.41 g
33.6 g
44.8 g
2

How many grams of \( \ce{CuO} \) are required to produce 8 g of \( \ce{Cu} \) with excess \( \ce{CO} \)? (Molar masses: \( \ce{CuO} \) = 79.5 g/mol, Cu = 64 g/mol)

Reaction: \( \ce{CuO + CO -> Cu + CO2} \).

Moles of Cu = \( \frac{8}{64} \) = 0.125 mol.

1 mol Cu from 1 mol \( \ce{CuO} \); mass = 0.125 × 79.5 ≈ 9.94 g.

4.97 g
15.9 g
12.5 g
9.94 g
4

A gas occupies 11.2 L at STP and weighs 16 g. What is its molar mass?

1 mole of gas at STP = 22.4 L.

Moles = \( \frac{11.2}{22.4} \) = 0.5 mol.

Molar mass = \( \frac{16}{0.5} \) = 32 g/mol.

16 g/mol
64 g/mol
32 g/mol
48 g/mol
3

How many grams of \( \ce{FeSO4} \) are required to produce 11.2 g of \( \ce{Fe} \) with excess \( \ce{H2} \)? (Molar masses: \( \ce{FeSO4} \) = 152 g/mol, Fe = 56 g/mol)

Reaction: \( \ce{FeSO4 + H2 -> Fe + H2SO4} \) (contextual reverse).

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

1 mol Fe from 1 mol \( \ce{FeSO4} \); mass = 0.2 × 152 = 30.4 g.

15.2 g
30.4 g
60.8 g
22.8 g
2

A 250 mL solution contains 4.8 g of \( \ce{NaCl} \) and has a density of 1.02 g/mL. What is the molarity? (Molar mass: \( \ce{NaCl} \) = 58.5 g/mol)

Moles of \( \ce{NaCl} \) = \( \frac{4.8}{58.5} \) ≈ 0.0821 mol.

Volume = 250 mL = 0.25 L.

Molarity = \( \frac{0.0821}{0.25} \) ≈ 0.328 M.

0.2 M
0.328 M
0.5 M
1.0 M
2

A mixture of 5 g \( \ce{H2} \) and 40 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: \( \ce{H2} \) = \( \frac{5}{2} \) = 2.5 mol, \( \ce{O2} \) = \( \frac{40}{32} \) = 1.25 mol.

2 mol \( \ce{H2} \) need 1 mol \( \ce{O2} \); 1.25 mol \( \ce{O2} \) limits, reacts with 2.5 mol \( \ce{H2} \) (available).

2.5 mol \( \ce{H2O} \) formed = 2.5 × 18 = 45 g.

45 g
36 g
54 g
18 g
1

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