Question

The equivalent conductivity of a solution containing $2.54g$ of $CuS{O}_4$ per litre is $91.0{\Omega }^{-1}c{m}^{2}e{q}^{-1}$. Its conductivity would be :

[Molar mass of $CuS{O}_4$ is $159.6g/mol$]

• A. $1.45\times {10}^{-3}{\Omega }^{-1}c{m}^{-1}$
• B. $2.17\times {10}^{-3}{\Omega }^{-1}c{m}^{-1}$
• C. $2.90\times {10}^{-3}{\Omega }^{-1}c{m}^{-1}$
• D. $3.9\times {10}^{-3}{\Omega }^{-1}c{m}^{-1}$

Answer 1

The correct option is C $2.90\times {10}^{-3}{\Omega }^{-1}c{m}^{-1}$

Given:
$\text{Mass,}m=2.54g$
$\text{Molar mass,}M=159.6g/mol$
$\text{Equivalent mass}=\frac{159.6}{2}\left(\text{n-factor is 2}\right)$
$\text{No. of gram equivalent}=\frac{\text{Mass}}{\text{Equivalent mass}}$

$\text{No. of gram equivalent}=\frac{2.54\times 2}{159.6}$

$\text{Normality}=0.032N$

The relationship between equivalence conductance and conductivity is ${\Lambda }_{eq}=\frac{1000\times \kappa }{N}\text{(when volume in}c{m}^3\text{)}$

Substitute values in the above expression,

$91=\frac{1000\times \kappa }{0.032}$

$\kappa =2.9\times {10}^{-3}{\Omega }^{-1}c{m}^{-1}$