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  4. Some Basic Concepts of Chemistry Chapter-Wise Test 18

Some Basic Concepts Of Chemistry Chapter-Wise Test 18

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A 0.56 g sample of a hydrocarbon produces 1.76 g of \( \ce{CO2} \) and 0.72 g of \( \ce{H2O} \) on complete combustion. What is its molecular formula if its molar mass is 56 g/mol? (Atomic masses: C = 12, H = 1, O = 16)

Mass of C = \( \frac{12}{44} \) × 1.76 ≈ 0.48 g; mass of H = \( \frac{2}{18} \) × 0.72 = 0.08 g.

Total = 0.48 + 0.08 = 0.56 g (matches).

Moles: C = \( \frac{0.48}{12} \) = 0.04, H = \( \frac{0.08}{1} \) = 0.08; ratio = 1 : 2; empirical formula = \( \ce{CH2} \), mass = 14 g/mol.

\( n = \frac{56}{14} = 4 \); molecular formula = \( \ce{C4H8} \).

\( \ce{C4H8} \)
\( \ce{C3H6} \)
\( \ce{C5H10} \)
\( \ce{C2H4} \)
1

What is the mass of \( \ce{AlCl3} \) produced when 13.5 g of \( \ce{Al} \) reacts with excess \( \ce{Cl2} \)? (Molar masses: Al = 27 g/mol, \( \ce{AlCl3} \) = 133.5 g/mol)

Reaction: \( \ce{2Al + 3Cl2 -> 2AlCl3} \).

Moles of Al = \( \frac{13.5}{27} \) = 0.5 mol.

2 mol Al produce 2 mol \( \ce{AlCl3} \); 0.5 mol produces 0.5 mol.

Mass = 0.5 × 133.5 = 66.75 g.

33.38 g
66.75 g
133.5 g
44.5 g
2

A 0.25 M \( \ce{NaOH} \) solution is diluted by adding 300 mL of water to 200 mL of the original solution. What is the final molarity?

\( M_1 V_1 = M_2 V_2 \).

0.25 × 200 = \( M_2 \) × (200 + 300).

\( M_2 = \frac{0.25 \times 200}{500} = 0.1 \) M.

0.15 M
0.05 M
0.2 M
0.1 M
4

A compound contains 52.17% carbon, 13.04% hydrogen, and 34.78% oxygen by mass. What is its empirical formula? (Atomic masses: C = 12, H = 1, O = 16)

For 100 g: C = 52.17 g, H = 13.04 g, O = 34.78 g.

Moles: C = \( \frac{52.17}{12} \) ≈ 4.35, H = \( \frac{13.04}{1} \) ≈ 13.04, O = \( \frac{34.78}{16} \) ≈ 2.17.

Ratio = 2 : 6 : 1; empirical formula = \( \ce{C2H6O} \).

\( \ce{CH3O} \)
\( \ce{C2H6O} \)
\( \ce{C2H5O} \)
\( \ce{CH2O} \)
2

How many significant figures are present in the result of \( \frac{0.0678 \times 310.15}{0.0821} \)?

0.0678 (3 sig figs), 310.15 (5 sig figs), 0.0821 (3 sig figs).

Result ≈ 256.29; limited to 3 sig figs (256).

2
4
3
5
3

What volume of \( \ce{O2} \) at STP is required to burn 6 g of \( \ce{CH4} \) completely to \( \ce{CO2} \) and \( \ce{H2O} \)? (Molar mass: \( \ce{CH4} \) = 16 g/mol)

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

Moles of \( \ce{CH4} \) = \( \frac{6}{16} \) = 0.375 mol.

1 mol \( \ce{CH4} \) needs 2 mol \( \ce{O2} \); 0.375 mol needs 0.75 mol.

Volume = 0.75 × 22.4 = 16.8 L.

16.8 L
8.4 L
22.4 L
33.6 L
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

How many grams of \( \ce{Al2O3} \) are required to produce 5.4 g of \( \ce{Al} \) with excess \( \ce{CO} \)? (Molar masses: \( \ce{Al2O3} \) = 102 g/mol, Al = 27 g/mol)

Reaction: \( \ce{Al2O3 + 3CO -> 2Al + 3CO2} \).

Moles of Al = \( \frac{5.4}{27} \) = 0.2 mol.

2 mol Al from 1 mol \( \ce{Al2O3} \); 0.2 mol from 0.1 mol; mass = 0.1 × 102 = 10.2 g.

5.1 g
20.4 g
15.3 g
10.2 g
4

What is the mass of \( \ce{MgSO4} \) produced when 8.4 g of \( \ce{Mg(OH)2} \) reacts with excess \( \ce{H2SO4} \)? (Molar masses: \( \ce{Mg(OH)2} \) = 58 g/mol, \( \ce{MgSO4} \) = 120 g/mol)

Reaction: \( \ce{Mg(OH)2 + H2SO4 -> MgSO4 + 2H2O} \).

Moles of \( \ce{Mg(OH)2} \) = \( \frac{8.4}{58} \) ≈ 0.1448 mol.

1 mol \( \ce{Mg(OH)2} \) produces 1 mol \( \ce{MgSO4} \); mass = 0.1448 × 120 ≈ 17.38 g.

8.69 g
17.38 g
34.76 g
12.0 g
2

A gas mixture contains 2 g of \( \ce{H2} \) and 16 g of \( \ce{O2} \). What is the total number of molecules? (Molar masses: \( \ce{H2} \) = 2 g/mol, \( \ce{O2} \) = 32 g/mol, Avogadro number = \( 6.022 \times 10^{23} \))

Moles of \( \ce{H2} \) = \( \frac{2}{2} \) = 1 mol; moles of \( \ce{O2} \) = \( \frac{16}{32} \) = 0.5 mol.

Total moles = 1 + 0.5 = 1.5 mol.

Molecules = 1.5 × \( 6.022 \times 10^{23} \) = \( 9.033 \times 10^{23} \).

\( 6.022 \times 10^{23} \)
\( 9.033 \times 10^{23} \)
\( 3.011 \times 10^{23} \)
\( 12.044 \times 10^{23} \)
2

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