Question

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

• A. $5 \times 10^2\, S \,cm^2\,mol^{-1}$
• B. $10^4\,S \,cm^2\,mol^{-1}$
• C. $200 \,S \,cm^2\,mol^{-1}$
• D. $0.02\, S \,cm^2\,mol^{-1}$

Answer 1

Answer: (D)

Molar conductivity, $\Lambda_{m}=\frac{\kappa\times1000}{m}$
M=molarity and $\kappa =$ specific conductivity
$\quad\quad \kappa=\frac{1}{R}\times$ cell constant
$\quad\quad=\frac{1}{R}\times\frac{l}{A}=\frac{1}{5\times10^{3} \Omega}\times\frac{1 cm}{100 cm^{2}}$
$\quad\quad =2 × 10^{-6} \Omega^{-1} cm^{-1}$
$\therefore \quad\Lambda_{m}=\frac{2\times10^{-6}\times1000}{0.1}=0.02 S cm^{2} mol^{-1}$