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Standard XII

Chemistry

Kohlrausch Law

A 0.05 M ...

Question

A $0.05 M$ $N H_{4} O H$ solution offers the resistance of $50 o h m$ to a conductivity cell at $298 K$ . If the cell constant is $0.50 {c m}^{- 1}$ and molar conductance of $N H_{4} O H$ at infinite dilution is $471.4 {o h m}^{- 1} {c m}^{2} {m o l}^{- 1}$ , calculate :

(i) Specific conductance
(ii) Molar conductance
(iii) Degree of dissociation

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Solution

(i) Specific conductance :
$\land_{s p} = \frac{1}{k}$

$= \frac{1}{50} = 0.02$

$\land_{s p} = 0.02 S {c m}^{- 1}$

(ii) Molar conductance :

$\land_{M}^{c} = \land_{s p} \times 1000 R$

$\land_{M}^{c} = 1000$

(iii) Degree of dissociation $(α)$

$α = \frac{\land_{M}^{c}}{\land_{M}^{\infty}}$

$α = \frac{1000}{471.4}$

$α = 2.12$

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Similar questions

The molar conductance of $N H_{4} O H$ at 0.01 M concentration is 11.3 $o h m^{- 1} c m^{2} m o l^{- 1}$ . The degree of dissociation of $N H_{4} O H$ is: (Molar conductance at infinite dilution = 271.1 $o h m^{- 1} c m^{2} m o l^{- 1}$

Q. The molar conductance of

N H_{4} O H

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o h m^{- 1} c m^{2} m o l^{- 1}

. The degree of dissociation
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Q. The specific conductance of a

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1.65 \times 10^{- 4}

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A 0.05M NH4OH solution offers the resistance of 50ohm to a conductivity cell at 298K. If the cell constant is 0.50cm−1 and molar conductance of NH4OH at infinite dilution is 471.4ohm−1cm2mol−1, calculate :(i) Specific conductance(ii) Molar conductance(iii) Degree of dissociation

(i) Specific conductance : ∧sp=1k =150=0.02 ∧sp=0.02Scm−1(ii) Molar conductance : ∧cM=∧sp×1000R ∧cM=1000(iii) Degree of dissociation (α) α=∧cM∧∞M α=1000471.4 α=2.12