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  4. Some Basic Concepts of Chemistry Chapter-Wise Test 15

Some Basic Concepts Of Chemistry Chapter-Wise Test 15

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A 0.4 M \( \ce{HCl} \) solution is diluted by adding 200 mL of water to 100 mL of the original solution. What is the final molarity?

\( M_1 V_1 = M_2 V_2 \).

0.4 × 100 = \( M_2 \) × (100 + 200).

\( M_2 = \frac{0.4 \times 100}{300} \) ≈ 0.133 M.

0.2 M
0.1 M
0.15 M
0.133 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 = \( \frac{52.17}{12} \approx 4.35 \), H = \( \frac{13.04}{1} \approx 13.04 \), O = \( \frac{34.78}{16} \approx 2.17 \) (3 significant figures).

Divide by smallest (2.17): C = \( \frac{4.35}{2.17} \approx 2.00 \), H = \( \frac{13.04}{2.17} \approx 6.00 \), O = \( \frac{2.17}{2.17} = 1.00 \).

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

A. \( \ce{CH3O} \)
B. \( \ce{C2H6O} \)
C. \( \ce{C4H12O2} \)
D. \( \ce{CH2O} \)
2

A mixture of 2.3 g ethanol (\( \ce{C2H5OH} \)) and 4.8 g methanol (\( \ce{CH3OH} \)) is dissolved in water to make 250 mL of solution. What is the total molarity of alcohols? (Molar masses: ethanol = 46 g/mol, methanol = 32 g/mol)

Moles of ethanol = \( \frac{2.3}{46} \) = 0.05 mol; moles of methanol = \( \frac{4.8}{32} \) = 0.15 mol.

Total moles = 0.05 + 0.15 = 0.2 mol.

Molarity = \( \frac{0.2}{0.25} \) = 0.8 M.

0.6 M
1.0 M
0.4 M
0.8 M
4

A 0.2 M \( \ce{NaCl} \) solution is diluted by adding 250 mL of water to 150 mL of the original solution. What is the final molarity?

\( M_1 V_1 = M_2 V_2 \).

0.2 × 150 = \( M_2 \) × (150 + 250).

\( M_2 = \frac{0.2 \times 150}{400} = 0.075 \) M.

0.1 M
0.05 M
0.15 M
0.075 M
4

What is the mass of \( \ce{MgO} \) produced when 12 g of \( \ce{Mg} \) reacts with 16 g of \( \ce{O2} \)? (Atomic masses: Mg = 24, O = 16)

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

Moles: Mg = \( \frac{12}{24} \) = 0.5 mol, \( \ce{O2} \) = \( \frac{16}{32} \) = 0.5 mol.

2 mol Mg need 1 mol \( \ce{O2} \); 0.5 mol \( \ce{O2} \) reacts with 1 mol Mg, but Mg limits at 0.5 mol.

0.5 mol \( \ce{MgO} \) formed = 0.5 × 40 = 20 g.

40 g
20 g
10 g
30 g
2

A mixture of 6 g \( \ce{H2} \) and 32 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, \( \ce{O2} \) = \( \frac{32}{32} \) = 1 mol.

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

2 mol \( \ce{H2O} \) formed = 2 × 18 = 36 g.

36 g
18 g
54 g
27 g
1

A compound contains 38.71% carbon, 9.68% hydrogen, and 51.61% oxygen by mass. What is its empirical formula? (Atomic masses: C = 12, H = 1, O = 16)

For 100 g: C = 38.71 g, H = 9.68 g, O = 51.61 g.

Moles: C = \( \frac{38.71}{12} \) ≈ 3.23, H = \( \frac{9.68}{1} \) ≈ 9.68, O = \( \frac{51.61}{16} \) ≈ 3.23.

Ratio = 1 : 3 : 1; empirical formula = \( \ce{CH3O} \).

\( \ce{CH2O} \)
\( \ce{CH3O} \)
\( \ce{C2H6O} \)
\( \ce{CH4O} \)
2

What is the mass of \( \ce{BaCl2} \) produced when 26.1 g of \( \ce{Ba(OH)2} \) reacts with excess \( \ce{HCl} \)? (Molar masses: \( \ce{Ba(OH)2} \) = 171 g/mol, \( \ce{BaCl2} \) = 208 g/mol)

Reaction: \( \ce{Ba(OH)2 + 2HCl -> BaCl2 + 2H2O} \).

Moles of \( \ce{Ba(OH)2} \) = \( \frac{26.1}{171} \) ≈ 0.1526 mol.

1 mol \( \ce{Ba(OH)2} \) produces 1 mol \( \ce{BaCl2} \); mass = 0.1526 × 208 ≈ 31.74 g.

15.87 g
47.61 g
20.8 g
31.74 g
4

How many significant figures are present in the result of \( \frac{0.0732 \times 310.15}{0.0821} \)?

0.0732 (3 sig figs), 310.15 (5 sig figs), 0.0821 (3 sig figs).

Result ≈ 276.47; limited to 3 sig figs (276).

2
4
3
5
3

How many significant figures are present in the result of \( \frac{0.0567 \times 298.15}{0.101} \)?

0.0567 (3 sig figs), 298.15 (5 sig figs), 0.101 (3 sig figs).

Result ≈ 167.34; limited to 3 sig figs (167).

2
4
3
5
3

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