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Some Basic Concepts Of Chemistry Chapter-Wise Test 1

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A 0.98 g sample of a hydrocarbon produces 3.08 g of \( \ce{CO2} \) and 1.26 g of \( \ce{H2O} \) on complete combustion. What is its molecular formula if its molar mass is 98 g/mol? (Atomic masses: C = 12, H = 1, O = 16)

Mass of C = \( \frac{12}{44} \) × 3.08 ≈ 0.84 g; mass of H = \( \frac{2}{18} \) × 1.26 = 0.14 g.

Total = 0.84 + 0.14 = 0.98 g (matches).

Moles: C = \( \frac{0.84}{12} \) = 0.07, H = \( \frac{0.14}{1} \) = 0.14; ratio = 1 : 2; empirical formula = \( \ce{CH2} \), mass = 14 g/mol.

\( n = \frac{98}{14} = 7 \); molecular formula = \( \ce{C7H14} \).

\( \ce{C7H14} \)
\( \ce{C6H12} \)
\( \ce{C5H10} \)
\( \ce{C8H16} \)
1

What is the mole fraction of \( \ce{C6H12O6} \) in a solution containing 18 g of \( \ce{C6H12O6} \) and 72 g of \( \ce{H2O} \)? (Molar masses: \( \ce{C6H12O6} \) = 180 g/mol, \( \ce{H2O} \) = 18 g/mol)

Moles of \( \ce{C6H12O6} \) = \( \frac{18}{180} \) = 0.1 mol; moles of \( \ce{H2O} \) = \( \frac{72}{18} \) = 4 mol.

Total moles = 0.1 + 4 = 4.1.

Mole fraction = \( \frac{0.1}{4.1} \) ≈ 0.0244.

0.05
0.0244
0.1
0.2
2

How many grams of \( \ce{O2} \) are required to react with 12 g of \( \ce{Mg} \) to form \( \ce{MgO} \) if the reaction proceeds with 75% efficiency? (Atomic masses: Mg = 24, O = 16)

Reaction: \( \ce{2Mg + O2 -> 2MgO} \).

Moles of Mg = \( \frac{12}{24} \) = 0.5 mol; needs 0.25 mol \( \ce{O2} \) (8 g).

With 75% efficiency, total \( \ce{O2} \) = \( \frac{8}{0.75} \) ≈ 10.67 g.

8 g
10.67 g
12 g
16 g
2

What is the molality of a solution prepared by dissolving 12 g of glucose (\( \ce{C6H12O6} \)) in 48 g of water, if the density of water is 1 g/mL? (Molar mass: \( \ce{C6H12O6} \) = 180 g/mol)

Moles of glucose = \( \frac{12}{180} \) ≈ 0.0667 mol.

Mass of water = 48 g = 0.048 kg.

Molality = \( \frac{0.0667}{0.048} \) ≈ 1.39 m.

1.39 m
0.69 m
2.78 m
1.00 m
1

A 300 mL solution contains 6 g of urea (\( \ce{NH2CONH2} \)) and has a density of 1.02 g/mL. What is its molarity? (Molar mass: urea = 60 g/mol)

Moles of urea = \( \frac{6}{60} \) = 0.1 mol.

Volume = 300 mL = 0.3 L.

Molarity = \( \frac{0.1}{0.3} \) ≈ 0.333 M.

0.2 M
0.333 M
0.5 M
1.0 M
2

A gas occupies 2.24 L at STP and weighs 3.2 g. What is its molar mass?

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

Molar mass = \( \frac{3.2}{0.1} \) = 32 g/mol.

16 g/mol
64 g/mol
32 g/mol
28 g/mol
3

A compound contains 36.36% carbon, 6.06% hydrogen, and 57.58% oxygen by mass. What is its empirical formula? (Atomic masses: C = 12, H = 1, O = 16)

For 100 g: C = 36.36 g, H = 6.06 g, O = 57.58 g.

Moles: C = \( \frac{36.36}{12} \) ≈ 3.03, H = \( \frac{6.06}{1} \) ≈ 6.06, O = \( \frac{57.58}{16} \) ≈ 3.6.

Ratio = 1 : 2 : 1.19 ≈ 1 : 2 : 1; empirical formula = \( \ce{CH2O} \).

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

A gas occupies 5.6 L at STP and weighs 2 g. What is its molar mass?

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

Molar mass = \( \frac{2}{0.25} \) = 8 g/mol.

4 g/mol
16 g/mol
8 g/mol
32 g/mol
3

What volume of \( \ce{O2} \) at STP is required to burn 9 g of \( \ce{C3H8} \) completely to \( \ce{CO2} \) and \( \ce{H2O} \)? (Molar mass: \( \ce{C3H8} \) = 44 g/mol)

Reaction: \( \ce{C3H8 + 5O2 -> 3CO2 + 4H2O} \).

Moles of \( \ce{C3H8} \) = \( \frac{9}{44} \) ≈ 0.2045 mol.

1 mol \( \ce{C3H8} \) needs 5 mol \( \ce{O2} \); 0.2045 mol needs 1.0225 mol.

Volume = 1.0225 × 22.4 ≈ 22.9 L.

22.9 L
11.2 L
33.6 L
4.58 L
1

What is the mass percentage of \( \ce{NaCl} \) in a solution made by dissolving 5.85 g of \( \ce{NaCl} \) in 44.15 g of water? (Molar mass: \( \ce{NaCl} \) = 58.5 g/mol)

Total mass = 5.85 + 44.15 = 50 g.

Mass % = \( \frac{5.85}{50} \) × 100 = 11.7%.

10%
15%
11.7%
12.5%
3

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