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Chemistry MCQs for NEET — Practice Questions with Answers

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Equivalent weight of a metal oxide is 20, the equivalent weight of sulphate of same metal will be

108

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The equivalent mass of H₃BO₃ (M = Molar mass of H₃BO₃) in its reaction with NaOH to from Na₂B₄O₇ is equal to –

$\frac{M}{2}$

$\frac{M}{4}$

$\frac{M}{6}$

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Explanation

H₃BO₃ is a mono basic acid

So $\frac{M}{1}$ = equivalent mass

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x gram of pure As₂S₃ is completely oxidised to respective highest oxidation states by 50 ml of 0.1 M hotacidified KMnO₄ then x mass of As₂S₃ taken is : (Molar mass of As₂S₃ = 246)

22.4 g

43.92 g

64.23 g

None

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Explanation

5As₂S₃ + 28KMnO₄ + H⁺→ 10H₃AsO₄ + 28Mn²⁺ + SO₄^2-

m moles of KMnO₄ = 50 × 0.1 = 5

28 mmoles of KMnO₄→ 5 mMoles of As₂S₃

1 mmoles of KMnO₄→ 5/28 mmoles of As₂S₃

Mass of As₂S₃ = x = 246 × $\frac{5}{28}$ g = 43.92 g

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The number of moles of ferrous oxalate oxidised by one mole of KMnO₄ is

$\frac{5}{2}$

$\frac{2}{5}$

$\frac{3}{5}$

$\frac{5}{3}$

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Explanation

Equivalents of FeC₂O₄ = equivalents of KMnO₄

x (mole) × 3 = 1 × 5

x = $\frac{5}{3}$

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In the reaction Na₂S₂O₃ + 4Cl₂ + 5H₂O → Na₂SO₄ + H₂SO₄ + 8HCI the equivalent weight of Na₂S₂O₃ will be

M/4

M/8

M/1

W/2

(M= molecular weight of Na₂S₂O₃)

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Explanation

Na₂ $\overset{+2}{s}$ ₂O₃→ Na₂ $\overset{+6}{s}$ O₄

see the total change in oxidation number = 4 × 2 = 8

∴ $E_{{Na}_{2} S_{2} O_{3}} = \frac{mol . wt .}{V . f} = \frac{M}{8}$

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An excess of NaOH was added to 100 mL of a ferric chloride solution. This caused the precipitation of 1.425 g of Fe(OH)₃. Calculate the normality of the ferric chloride solution

0.20 N

0.50 N

0.25 N

0.40 N

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Explanation

3 NaOH + FeCl₃ → Fe(OH)₃ + 3NaCl

m.e. of NaOH = me.e of Fe(OH)₃

100 × N = $\frac{W}{E} \times 1000 (E_{{Fe(OH)}_{3}} = \frac{mol. wt.}{3})$

N = $\frac{1 .425 \times 10 \times 3}{107}$ = 0.3999 = 0.4 N

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In the reaction CrO₅ + H₂SO₄→ Cr₂(SO₄)₃ + H₂O + O₂ one mole of CrO₅ will liberate how many moles of O₂

5/2

5/4

9/2

none of these

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Explanation

The balance reaction is

4CrO₅ + 2H₂SO₄ → 2 Cr₂ (SO₄)₃ + 2H₂O + O₂

∴ 1 mole CrO5 can liberate only $\frac{1}{4}$ mole O₂

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0.4g of a polybasic acid H_nA (all the hydrogens are acidic) requires 0.5g of NaOH for complete neutralization. The number of replaceable hydrogen atoms in the acid and the molecular weight of 'A' would be : (Molecular weight of the acid is 96 gms.)

1, 95

2, 94

3, 93

4, 92

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Explanation

Eq of Acid = Eq of base

n × $\frac{0 .4}{96} = \frac{0 .5}{40}$

n = $\frac{1}{80} \times \frac{96}{0 .4} = \frac{96}{32} = 3$

∴ wt of A = 96 – 3 = 93.

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25.0 g of FeSO₄.7H₂O was dissolved in water containing dilute H₂SO₄, and the volume was made up to 1.0 L. 25.0 mL of this solution required 20 mL of an N/10 KMnO₄ solution for complete oxidation. The percentage of FeSO₄^. 7H₂O in the acid solution is

78%

98%

89%

79%

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Explanation

M.e. of FeSO₄⋅ 7H₂O in 25 ml = m.e. of KMnO₄ used = 2 m.e.

M.e. of FeSO₄⋅ 7H₂O in 1000 ml = 80 m.e.

mass of FeSO₄⋅ 7H₂O in solution = $\frac{80}{1} \times 278.1 \times \frac{1}{1000} = 22 .24 gm$

% of FeSO₄ ⋅ 7H₂O = $\frac{22 .24}{25}$ × 100 = 88.96 89%

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25 mL of a solution containing HCl and H₂SO₄ required 10 mL of a 1 N NaOH solution for neutralization. 20 mL of the same acid mixture on being treated with an excess of AgNO₃ gives 0.1425 g of AgCf. The normality of the HC and the normality of the H₂SO₄ are respectively

0.40 N and 0.05 N

0.05 N and 0.35 N

0.50 N and 0.25 N

0.40 N and 0.5 N

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Explanation

Let normality of HCl is N₁ and H₂SO₄ is N₂.

∴ M.e. of HCl + M.e. of H₂SO₄ = M.e of NaOH

25 × N₁ + 25 × N₂ = 10 × 1 ... (1)

N₁ + N₂ = 0.4 ... (1)

BY POAC

Moles of Cl = moles of AgCl

$\frac{20 \times N_{1}}{1000} = \frac{0 .1435}{143 .05} = 10^{- 3}$

N₁ = 0.05 N

N₂ = 0.35 N

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