Resistance of 0.2 M solution of an electrolyte is $50\mathrm{\Omega }$. The specific conductance of the solution is 1.3 Sm^-1.If resistance of the 0.4 M solution of the same electrolyte is $260\mathrm{\Omega }$,its molar conductivity is

Specific conductance = conductance x cell constant 
             $1.3{\mathrm{Sm}}^{-1}=\frac{1}{50}\mathrm{S}\times \mathrm{cell} \mathrm{constant}$
$\therefore \text{ Cell constant }=1.3\times 50{\mathrm{m}}^{-1}=65 {\mathrm{m}}^{-1}=(65/100){\mathrm{cm}}^{-1}$
$\text{ Molar conductivity =}\frac{1000\times \mathrm{conductance} \times \mathrm{cell} \mathrm{constant} }{ \mathrm{molarity} }$
                                 $\begin{array}{l}=\frac{1000}{0.4}\times \frac{1}{260}\times \frac{65}{100}=6.25{\mathrm{Scm}}^2{\mathrm{mol}}^{-1}\\ =6.25\times {10}^{-4}{\mathrm{Sm}}^2{\mathrm{mol}}^{-1}\end{array}$