Question

$0.1M$ solution is present in a conductivity cell with electrode of $100c{m}^2$ area placed 1 cm apart and resistance observed is $5\times 10^3$ ohm. What is the molar conductivity of this solution?

• A. $5\times 10^{2}Sc{m}^{2}mo{l}^{-1}$
• B. $10^{4}Sc{m}^{2}mo{l}^{-1}$
• C. $200Sc{m}^{2}mo{l}^{-1}$
• D. $0.02Sc{m}^{2}mo{l}^{-1}$

Answer 1

Answer: (D)

Molar conductivity, ${\Lambda }_m=\frac{\kappa \times 1000}{m}$
M=molarity and $\kappa =$ specific conductivity
$\kappa =\frac{1}{R}\times$ cell constant
$=\frac{1}{R}\times \frac{l}{A}=\frac{1}{5\times 10^3\Omega }\times \frac{1cm}{100c{m}^2}$
$=2\times 10^{-6}{\Omega }^{-1}c{m}^{-1}$
$\therefore {\Lambda }_m=\frac{2\times 10^{-6}\times 1000}{0.1}=0.02Sc{m}^{2}mo{l}^{-1}$