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Chemistry NEET Questions — Free MCQ Practice

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The specific conductance has the unit :

A.

$o h m^{- 1} c m^{- 1}$

B.

ohm.cm

C.

$o h m c m^{- 1}$

D.

$o h m^{- 1} . c m$

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Explanation

(A). Specific conductance = Observed conductance $\times \frac{l}{a}$

$= o h m^{- 1} \times \frac{c m}{c m^{2}} = o h m^{- 1} c m^{- 1}$

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What is the value of

$p K_{b} (C H_{3} C O O^{-}) i f λ_{m}^{0} = 390 & λ_{m} = 7.8 f o r 0.04 o f a C H_{3} C O O H a t 25 ° C$

A.

9.3

B.

9.2

C.

4.7

D.

4.8

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Explanation

$α = \frac{λm}{λ^{0} m} = \frac{7.8}{390} = 0.02 {Cα}^{2} = Ka = 0.047 \times {(0.02)}^{2} = 16 \times 10^{- 6} pKa = - \log Ka = - \log 16 \times 10^{- 6} 6 - \log {(2)}^{4} = 6 - 4 \times 0.3 = 4.8 pKa = 14 - pKa = 14 - 4.8 = 9.2$

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The number of faradays required to produce one mole of water from a hydrogen - oxygen fuel cell containing aqueous alkali as electrolyte is-

A.

1

B.

3

C.

2

D.

4

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Explanation

C). This is based on the application of Faraday’s laws of electrolysis. One mole of water can be obtained by the combination of a mole of $H_{2}$ and $\frac{1}{2}$ mole of oxygen which may be produced by the passage of 2 Faradays.

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How much current is necessary to produce $H_{2}$ gas at the rate of 1 $c m^{3}$ per second under STP.

A.

4.305 amp

B.

17.22 amp

C.

8.61 amp

D.

2.1525 amp.

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Explanation

(C). 1 mol H₂ gas at STP = $2 F = 2 \times 96500 C$

$22400 c m^{3} o f H_{2} g a s a t S T P = 193 \times 10^{3} C$

$1 c m^{3} o f H_{2}$ gas at STP per second $= \frac{193 \times 10^{3}}{22400} = 8.61 a m p .$

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A current of 0.250 A is passed through 400 ml of a 2.0 M solution of NaCl for 35 minutes. What will be the pH of the solution after the current is turned off ?

A.

12.98

B.

12.13

C.

10.48

D.

9.24

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Explanation

(B). After electrolysis aqueous NaCl is converted into aqueous NaOH.
The quantity of electricity passed

$= \frac{0.250 \times 35 \times 60}{96500} F = 5.44 \times 10^{- 3} F$

The number of equivalents of $O H^{-}$ ion formed

$= 5.44 \times 10^{- 3}$

$∴$ Molarity of NaOH = $\frac{5.44 \times 10^{- 3}}{0.4 L} = 1.36 \times 10^{- 2}$

∴ pOH = – log $(1.36 \times 10^{- 2}) = 1.87$

$∴$ pH = 12.13

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During discharge of a lead storage cell the density of sulphuric acid in the cell-

A.

Increases

B.

Decreases

C.

Remains unchanged

D.

Initially increases but decreases subsequently

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Explanation

(B). During the discharge of lead storage cell, sulphuric acid is consumed. Its concentration decreases and therefore, density decreases.

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The standard electrode potentials (reduction) of $P t / F e^{2 +}, F e^{3 +} a n d P t / S n^{4 +}, S n^{2 +} a r e + 0.77 V a n d + 0.15 V$ respectively at $25^{0} C .$ The standard EMF of the reaction $S n^{4 +} + 2 F e^{2 +} \to S n^{2 +} + 2 F e^{3 +} i s -$

A.

– 0.62 V

B.

– 0.92 V

C.

+ 0.31 V

D.

+ 0.85 V

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Explanation

(A). For the given reaction

$E = E_{S n^{2 +} | S n^{4 +}}^{°} - E_{F e^{2 +}, F e^{3 +} =}^{°} 0.15 - 0.77 = - 0.62 V$

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Specific conductance of 0.01 M KCl solution is x $o h m^{- 1} c m^{- 1}$ . When conductivity cell is filled with 0.01 M KCl the conductance observed is y $o h m^{- 1}$ . When the same cell is filled with 0.01 M $H_{2} S O_{4}$ the observed conductance was $z o h m^{- 1} c m^{- 1} .$ Hence specific conductance of 0.01 M $H_{2} S O_{4}$ is-

A.

xz

B.

z/xy

C.

xz/y

D.

xy/z

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Explanation

C). Cell constant

$= \frac{S p e c i f i c c o n d u c \tan c e}{O b s e r v e d c o n d u c \tan c e} = \frac{x}{y} c m^{- 1}$

Specific conductance of $0.01 M H_{2} S O_{4}$

= Observed conductance $\times$ Cell constant $= z \times \frac{x}{y} o h m^{- 1} c m^{- 1}$

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$F_{2}$ gas can’t be obtained by the electrolysis of any $F^{- 1}$ salt because-

A.

Fluorine is the strongest reducing agent

B.

Fluorine is the strongest oxidising agent.

C.

Fluorine easily combine with atmospheric $O_{2}$

D.

All

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Explanation

(B). Since Fluorine is the strongest oxidising agent so it will destroy the electrode employed.

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At 298K the standard free energy of formation of $H_{2} O (l) i s - 237.20 k J / m o l e$ while that of its ionisation into $H^{+}$ ion and hydroxyl ions is 80 kJ/mol. then the emf of the following cell at 298 K will be (take F = 96500 C]

$H_{2} (g, 1 b a r) H^{+} (1 M) | | O H^{-} (1 M) | O_{2} (g, 1 b a r)$

A.

0.40 V

B.

0.81 V

C.

1.23 V

D.

–0.40 V

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Explanation

(A) Cell reaction

Cathode : $H_{2} O (l) + \frac{1}{2} O_{2} (g) + 2 e^{-} \to 2 O H^{-} (a q)$

Anode : $H_{2} (g) \to 2 H^{+} (a q .) + 2 e^{-}$

$H_{2} O (l) + \frac{1}{2} O_{2} (g) + H_{2} (g) \to 2 H^{+} (a q .) + 2 O H^{-} (a q .)$

Also we have

$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l) . . . . . . . (1) ∆ G °_{f} = - 237.2 k J / m o l H_{2} O (l) \to H^{+} (a q .) + O H^{-} (a q .) . . . . . . . (2) ∆ G °_{f} = - 237.2 k J / m o l$

Hence for cell reaction $[e q . (1) + e q (2) \times 2]$

$∆ G ° = - 77.20 k J / m o l S o, E ° - \frac{∆ G °}{n F} = \frac{77200}{2 \times 96500} = 0.40 V$

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