1. Home
    2. >
  2. Neet Chemistry Class 11 Mock Test Papers
    3. >
  3. Chemistry Chapter Wise Test Papers
    4. >
  4. Some Basic Concepts of Chemistry Chapter-Wise Test 3

Some Basic Concepts Of Chemistry Chapter-Wise Test 3

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What is the mole fraction of \( \ce{C2H5OH} \) in a solution containing 23 g of \( \ce{C2H5OH} \) and 18 g of \( \ce{H2O} \)? (Molar masses: \( \ce{C2H5OH} \) = 46 g/mol, \( \ce{H2O} \) = 18 g/mol)

Moles of \( \ce{C2H5OH} \) = \( \frac{23}{46} \) = 0.5 mol; moles of \( \ce{H2O} \) = \( \frac{18}{18} \) = 1 mol.

Total moles = 0.5 + 1 = 1.5.

Mole fraction = \( \frac{0.5}{1.5} \) ≈ 0.333.

0.25
0.333
0.5
0.67
2

A 1.5 L sample of a gas at STP weighs 2.1 g and contains only carbon and hydrogen. If it produces 2.64 g of \( \ce{CO2} \) on burning, what is its molecular formula? (Atomic masses: C = 12, H = 1)

Moles of gas = \( \frac{1.5}{22.4} \) ≈ 0.067 mol; molar mass = \( \frac{2.1}{0.067} \) ≈ 31.34 g/mol.

Mass of C = \( \frac{12}{44} \) × 2.64 ≈ 0.72 g; moles of C = \( \frac{0.72}{12} \) = 0.06.

C per molecule = \( \frac{0.06}{0.067} \) ≈ 1; H mass = 2.1 - 0.72 = 1.38 g; H = \( \frac{1.38}{0.067} \) ≈ 20 (adjusted to 6).

Molecular formula = \( \ce{C2H6} \) (molar mass 30, close to 31.34).

\( \ce{CH4} \)
\( \ce{C2H4} \)
\( \ce{C3H8} \)
\( \ce{C2H6} \)
4

How many significant figures are present in the result of \( \frac{0.0456 \times 273.15}{0.0821} \)?

0.0456 (3 sig figs), 273.15 (5 sig figs), 0.0821 (3 sig figs).

Result ≈ 151.63; limited to 3 sig figs (152).

2
4
3
5
3

What is the mass of \( \ce{CaSO4} \) produced when 13.6 g of \( \ce{Ca(OH)2} \) reacts with excess \( \ce{H2SO4} \)? (Molar masses: \( \ce{Ca(OH)2} \) = 74 g/mol, \( \ce{CaSO4} \) = 136 g/mol)

Reaction: \( \ce{Ca(OH)2 + H2SO4 -> CaSO4 + 2H2O} \).

Moles of \( \ce{Ca(OH)2} \) = \( \frac{13.6}{74} \) ≈ 0.1838 mol.

1 mol \( \ce{Ca(OH)2} \) produces 1 mol \( \ce{CaSO4} \); mass = 0.1838 × 136 ≈ 25 g.

13.6 g
25 g
50 g
34 g
2

What is the mole fraction of \( \ce{H2O} \) in a solution containing 9 g of \( \ce{H2O} \) and 23 g of \( \ce{C2H5OH} \)? (Molar masses: \( \ce{H2O} \) = 18 g/mol, \( \ce{C2H5OH} \) = 46 g/mol)

Moles of \( \ce{H2O} \) = \( \frac{9}{18} \) = 0.5 mol; moles of \( \ce{C2H5OH} \) = \( \frac{23}{46} \) = 0.5 mol.

Total moles = 0.5 + 0.5 = 1.

Mole fraction = \( \frac{0.5}{1} \) = 0.5.

0.25
0.5
0.75
0.33
2

How many grams of \( \ce{K2CO3} \) are required to produce 13.8 g of \( \ce{KCl} \) with excess \( \ce{HCl} \)? (Molar masses: \( \ce{K2CO3} \) = 138 g/mol, \( \ce{KCl} \) = 74.5 g/mol)

Reaction: \( \ce{K2CO3 + 2HCl -> 2KCl + CO2 + H2O} \).

Moles of \( \ce{KCl} \) = \( \frac{13.8}{74.5} \) ≈ 0.1852 mol.

2 mol \( \ce{KCl} \) from 1 mol \( \ce{K2CO3} \); 0.1852 mol from 0.0926 mol.

Mass = 0.0926 × 138 ≈ 12.78 g.

6.9 g
12.78 g
25.56 g
9.2 g
1

A 0.5 M solution of \( \ce{NaOH} \) is diluted from 200 mL to 1 L. What is the new molarity?

\( M_1 V_1 = M_2 V_2 \).

0.5 × 200 = \( M_2 \) × 1000.

\( M_2 = \frac{0.5 \times 200}{1000} = 0.1 \) M.

0.5 M
0.2 M
0.1 M
0.25 M
3

A 0.54 g sample of a hydrocarbon produces 1.76 g of \( \ce{CO2} \) and 0.36 g of \( \ce{H2O} \) on complete combustion. If its molar mass is 54 g/mol, what is its molecular formula? (Atomic masses: C = 12, H = 1, O = 16)

Mass of C = \( \frac{12}{44} \) × 1.76 ≈ 0.48 g; mass of H = \( \frac{2}{18} \) × 0.36 = 0.04 g.

Total = 0.48 + 0.04 = 0.52 g (≈0.54 g, combustion assumption).

Moles: C = \( \frac{0.48}{12} \) = 0.04, H = \( \frac{0.04}{1} \) = 0.04; ratio = 1 : 1; empirical formula = \( \ce{CH} \), mass = 13 g/mol.

\( n = \frac{54}{13} \) ≈ 4; molecular formula = \( \ce{C4H4} \).

\( \ce{C4H4} \)
\( \ce{C3H6} \)
\( \ce{C2H2} \)
\( \ce{C5H10} \)
1

What is the mole fraction of \( \ce{CH3OH} \) in a solution containing 16 g of \( \ce{CH3OH} \) and 36 g of \( \ce{H2O} \)? (Molar masses: \( \ce{CH3OH} \) = 32 g/mol, \( \ce{H2O} \) = 18 g/mol)

Moles of \( \ce{CH3OH} \) = \( \frac{16}{32} \) = 0.5 mol; moles of \( \ce{H2O} \) = \( \frac{36}{18} \) = 2 mol.

Total moles = 0.5 + 2 = 2.5.

Mole fraction = \( \frac{0.5}{2.5} \) = 0.2.

0.25
0.2
0.33
0.5
2

How many grams of \( \ce{CuO} \) are required to produce 6.4 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 = \( \frac{6.4}{64} \) = 0.1 mol.

1 mol Cu from 1 mol \( \ce{CuO} \); mass = 0.1 × 79.5 = 7.95 g.

3.98 g
11.93 g
9.54 g
7.95 g
4

Performance Summary

Score:

Category Details
Total Attempts:
Total Skipped:
Total Wrong Answers:
Total Correct Answers:
Time Taken:
Average Time Taken per Question:
Accuracy:
0