Question

The resistance and conductivity of 0.02 M KCl solution are 82.4 ohm and $0.002768Sc{m}^{-1}$, respectively. When filled with $0.005N{K}_{2}S{O}_4$, the solution had a resistance of 324 ohm. Calculate equivalent conductivity of $K_{2}S{O}_4$ solution.

• A. $381.6Sc{m}^{2}mo{l}^{-1}$
• B. $581.6Sc{m}^{2}mo{l}^{-1}$
• C. $681.6Sc{m}^{2}mo{l}^{-1}$
• D. $281.6Sc{m}^{2}mo{l}^{-1}$

Answer 1

The correct option is D $281.6Sc{m}^{2}mo{l}^{-1}$

${\wedge }_{eq}=\frac{k\times 1000}{N}=\frac{7.04\times {10}^{-4}\times 1000}{\frac{0.005}{2}}$
$=281.6Sc{m}^{2}mo{l}^{-1}$
[n factor for $K_{2}S{O}_4=2;M=\frac{N}{2}$]