Question

The equivalent conductivity of $H_{2}S{O}_4$ solution is $1.3\times {10}^2{\text{ohm}}^{-1}{\text{cm}}^2{\text{eq}}^{-1}$. If its specific conductance is $26\times {10}^{-2}{\text{ohm}}^{-1}{\text{cm}}^{-1}$, then concentration of $H_{2}S{O}_4$ solution is :

• A. $2\text{M}$
• B. $1\text{M}$
• C. $0.2\text{M}$
• D. $0.1\text{M}$

Answer 1

The correct option is B $1\text{M}$

$\text{Given : Equivalent conductivity}\left({\Lambda }_{eq}\right)=1.3\times {10}^2{\text{ohm}}^{-1}{\text{cm}}^2{\text{eq}}^{-1}$

$\text{Specific conductance (K)}=26\times {10}^{-2}{\text{Ohm}}^{-1}{\text{cm}}^{-1}$
We know,
${\Lambda }_{eq}=\frac{1000\times K}{N}$
$\Rightarrow 1.3\times {10}^2=\frac{1000\times 26\times {10}^{-2}}{M\times 2}(\text{Normality =Molarity}\times \text{n-factor})$
$\Rightarrow M=\frac{1000\times 26\times {10}^{-2}}{1.3\times {10}^2\times 2}(\text{n-factors of}{H}_{2}S{O}_4=2)$

$\Rightarrow \text{Molarity = 1 M}$