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Chemistry MCQs for NEET — Practice Questions with Answers

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If 10 gm of V₂O₅ is dissolved in acid and is reduced to V²⁺ by zinc metal, how many mole of I₂ could be reduced by the resulting solution if it is further oxidised to VO²⁺ions ? [Assume no change in state of Zn²+ions] (V = 51, O =16, I = 127) :

0.11 mole of I₂

0.22 mole of I₂

0.055 mole of I₂

0.44 mole of I₂

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Explanation

6e^- + 10H⁺ + V₂O₅ → 2V2⁺ + 5 H₂ O

Zn → Zn²⁺ + 2e^- 3 × 3

V₂O₅ + 3 Zn + 10 H⁺→ 3Zn²⁺ + 2V²⁺ + 5H₂O …

Now H₂O + V²⁺→ VO²⁺ + 2H⁺ + 2e^-

2e^- + I₂→ 2r^-

V²⁺ + I₂ + H₂O → 2I^- + VO²⁺ + 2H⁺

so we have 1 moles of V₂O₅ will reduce 2 moles of iodine

so $(\frac{10}{102 + 80}) × 2$ moles of will be reduced by given amount of V₂O₅ = 0.11moles of 1₂

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0.70 g of mixture (NH₄)₂ SO₄ was boiled with 100 mL of 0.2 N NaOH solution till all the NH₃(g) evolved and get dissolved in solution itself. The remaining solution was diluted to 250 mL. 25 mL of this solution was neutralized using 10 mL of a 0.1 N H₂SO₄ solution. The percentage purity of the (NH₄)₂ SO sample is

94.3

50.8

47.4

79.8

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Explanation

M.e. of NH₃ = 10

m.e. of NH3 = 10

m.mole of (NH₄)₂SO₄= 5

wt. of (NH₄)₂ SO₄ = 5

wt. of (NH4)2 SO4 = $\frac{5}{1000} \times$ 132 = 0.66 gm

% of (NH₄)₂ S0₄= $\frac{0 .66}{0 .7} \times$ 100 = 94.28 %

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A mixed solution of potassium hydroxide and sodium carbonate required 15 mL. of an N/20 HCI solution when titrated with phenolphthalein as an indicator. But the same amount of the solution when titrated with methyl orange as an indicator required 25 mL of the same acid. The amount of KOH present in the solution is

0.014g

0.14g

0.028g

1.4 g

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Explanation

KOH + Na₂aO₃

a M.e.b.M.e.

a + $\frac{b}{2}$ = 15 × $\frac{1}{20}$

2a + b= 1.5 … (i) (in presence of phenolphthalein)

a + b = 25 × $\frac{1}{20}$ = 1.25… (ii) (in presence of Methyl orange)

by solving (i) & (ii) a = 0.25 m.e

mass of KOH = $\frac{0 .25}{1000} \times$ 56 = 0. 014 gm

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When a solution containing 4.77 gm. of NaCl is added to a solution of 5.77 gm. of AgNO₃, the weight of precipitated AgCl is [IIT 1978]

11.70 gm.

9.70 gm.

4.86 gm.

2.86 gm.

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Explanation

AgNO₃ + NaCl → AgCl + NaNO₃

Moles before mixing $\frac{5 .77}{108 + 14 + 48}$ $\frac{4 .77}{58 .5}$ 0 0

= 0.0339 = 0.0815 (Here AgNO₃ is limiting reactant, thus)

Moles after mixing 0 0.0815 – 0.0339 0.0339 0.0339

= 0.0476

∴ Moles of AgCl formed = 0.0339

∴ Mass of AgCl formed = Mol. mass × No. of moles = 143.5 × 0.0339 = 4.864 gm

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In which of the following compounds, nitrogen exhibits highest oxidation state?

N₂H₄

NH₃

N₃H

NH₂OH

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Explanation

Let the oxidation state of nitrogen in the given compounds be x.

(a) x +1

N₂H₄

2x + (+1)4 = 0

2x = -4

.^.. x = -2

(b) x +1

NH₃ x + (+1)3 =0

.^.. x = -3

N₃H x3 + (+1)0 = 0

3x = -1

.^.. x = -1/3

(d) x+1-2+1

NH₂OH x + (+1)2+(-2)+(+1) =0

x+2-2+1=0

.^.. x =-1

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A mixture of potassium chlorate, oxalic add and sulphuric acid is heated. During the reaction which element undergoes a maximum change in the oxidation number?

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Oxidation states of P in H₄P₂O₅, H₄P₂O₆, H₄P₂O₇, arc respectively

+3, +5, +4

+5, +3, +4

+5, +4, +3

+3, +4, +5

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Explanation

Key Idea Oxidation state of H is +1 and that of O is -2.

Let the oxidation state of P in the given compounds is x.

In H₄P₂O₅ .

(+1) x 4 + 2 x x + (- 2) x 5 = 0

4+ 2x-10 = 0

2x = 6

x = + 3

In H₄P₂O₆,

(+1)x4+2x + (-2)x6 = 0

4 + 2x-12=0

2x = 8

x = + 4

In H₄P₂O₇,

(+1)x4+2 x (x + (-2)x7)=0

4+2x -14 = 0

2x=10

x=+ 5

Thus, the oxidation states of P in H₄P₂O₅, H₄P₂O₆, H₄P₂O₇ are + 3 + 4 and +5 respectively.

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Oxidation numbers of P in ${PO}_{4}^{3 -}$ , of S in ${SO}_{4}^{2 -}$ and that of Cr in ${Cr}_{2} O_{7}^{2 -}$ are respectively,

+5, +6 and +6

+3, +6 and +5

+5, +3 and +6

-3, +6 and +6

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Explanation

Key Idea (i) Sum of oxidation states of all atoms = charge of ion.

(ii) Oxidation number of oxygen = -2

Let the oxidation state of P in ${PO}_{4}^{3 -}$ be x.

${PO}_{4}^{3 -}$

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Number of moles of MnO^-₄ required to oxidise one mole of ferrous oxalate completely in acidic medium will be

0.6 mole

0.4 mole

7.5 moles

0.2 mole

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The number of mole of KMnO₄ that will be needed to react with one mole of sulphite ion in acidic solution is:

$\frac{3}{5}$

$\frac{4}{5}$

$\frac{2}{5}$

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