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NEET NEET Questions — Free MCQ Practice

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A 2 gm mixture of K₂CO₃ and KCl is completely reacted with 100 ml N/10 HCl. The % of K₂CO₃ in the mixture is (K = 39, Cl =35.5, C=12, O=16)

A.

69%

B.

34.5%

C.

50%

D.

100%

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Explanation

2 gm mixture of Kâ‚‚COâ‚ƒ and KCl is treated with 100 ml N/10 HCl. Kâ‚‚COâ‚ƒ neutralizes 2 equivalents of HCl while KCl remains unreacted. Using the mole concept, it can be calculated that the % of Kâ‚‚COâ‚ƒ in the mixture is 34.5%.

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400 ml M/10 H₂SO₄ is mixed with 600 ml N/10 NaOH then normality and nature of solution respectively will be

A.

0.02N, basic

B.

0.02N, acidic

C.

0.04N, acidic

D.

0.14N , acidic

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Explanation

When 400 ml M/10 Hâ‚‚SOâ‚„ (0.4 mol) is mixed with 600 ml N/10 NaOH (0.6 mol), the excess NaOH neutralizes some Hâ‚‚SOâ‚„, leaving the solution slightly acidic with a normality of 0.02N.

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The oxidation number of sulphur in S₈, S₂F₂ and H₂S respectively are :

A.

0, +1 and -2

B.

+2, +1 and -2

C.

0, +1 and +2

D.

-2. +1 and -2

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Explanation

In S₈, S₂F₂ and H₂S, the oxidation number of S is 0, +1 and -2 respectively.

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When Cl₂ is converted into Cl^- & Cl $O_{3}^{-}$ then n-factor of Cl₂will be:

A.

3/5

B.

5/3

C.

7/2

D.

2

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Explanation

When Cl2 is converted to Cl- and ClO3-, the change in oxidation state is from 0 to -1 and +5 respectively. The overall change is 5 units. Therefore, the n-factor for Cl2 is 5/3 according to the formula n = (highest O.S. - lowest O.S.)/number of atoms undergoing change.

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MnO₂ + 4HCl $\to$ MnCl₂+2H₂O+Cl₂, the equivalent wt. of HCl will be (M $\equiv$ Mol wt of HCl)

A.

M

B.

M/2

C.

2M

D.

M/4

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Which acts as a reducing agent only?

A.

SO₂

B.

HNO₃

C.

MnO₂

D.

H₂S

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Explanation

IN H₂S, S has lowest oxidation number i.e. -2

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Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox titration. Some half cell reactions and their standard potential are given below

Mn $O_{4}^{-}$ (aq) + 8H⁺(aq) + 5e^- $\to$ Mn²⁺(aq) + 4H₂O(l); $E^{°}$ =1051V

Cr₂O_7^2-(aq) +14H⁺(aq) + 6e _$\to$2Cr_³⁺(aq)₊7H₂O(l)_{; $E^{°}$}=1.38V

Fe³⁺(aq) + e^- _$\to$Fe²⁺(aq);_$E^{°}$=0.77V

Cl₂(g) + 2e^- _$\to$2Cl^-(aq); _$E^{°}$= 1.40V

Identify the only incorrect statement regarding the quantitative estimation of aqueous Fe(NO₃)₂

A.

Mn $O_{4}^{-}$ can be used in aqueous HCl

B.

Cr₂O_7^2-can be used in aqueous HCl

C.

Mn $M_{4}^{2 -}$ can be used in aqueous H₂SO₄

D.

Cr₂O_7^2-can be used in aqueous H₂SO₄

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Explanation

From reduction electrode potential values, Mn $O_{4}^{-}$ can oxidise Cl^- as well as Fe²⁺. Hence, Mn $O_{4}^{-}$ cannot be used in aqueous HCl for quantitative estimation of Fe(NO₃)₂.

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The equivalent weight of H₃PO₂, when it disproportionates into PH₃ and H₃PO₃ is

A.

66

B.

49.5

C.

33

D.

20.5

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Explanation

The disproportionation reaction of H3PO2 is: 2H3PO2 â†’ PH3 + H3PO3. In this reaction, 2 moles of H3PO2 produce 1 mole of PH3 and 1 mole of H3PO3. Therefore, the equivalent weight of H3PO2 is (2 Ã— molecular weight of H3PO2) / 2 = 49.5.

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In the reaction, 2Cu⁺ $\to$ Cu + Cu²⁺, the equivalent weight of Cu⁺ is (M is the mol. wt. of Cu⁺)

A.

M

B.

M/2

C.

M/4

D.

2M

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Explanation

2Cu⁺ $\to$ Cu + Cu²⁺

Cu⁺ $\to$ Cu M

Cu⁺ $\to$ Cu²⁺, M

Equivalent weight of Cu⁺ M+M=2M

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Equal volumes of 1M each of KMnO₄ and $K_{2} {Cr}_{2} O_{7}$ are used to oxidize Fe(II) solution in acidic medium. The amount of Fe oxidized will be

A.

More with KMnO₄

B.

More with K₂Cr₂ $O_{7}$

C.

Equal with both oxidising agent

D.

Cannot be determined

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Explanation

1M KMnO₄ = 5 equivalents KMnO₄

1M K₂Cr₂ $O_{7}$ = 6 equivalents K₂Cr₂ $O_{7}$

So, K₂Cr₂ $O_{7}$ oxidises more Fe²⁺ in comparision to KMnO₄

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