Question

The resistance of 0.05 $ M\, CH_3COOH$ solution is found to be 100 ohm. If the cell constant is $0.037 \,cm^{-1}$ the molar conductivity of the solution is

• A. $3.7 ohm^{-1} cm^2 mol^{-1}$
• B. $74 ohm^{-1} cm^2 mol^{-1}$
• C. $7.4 ohm^{-1} cm^2 mol^{-1}$
• D. $37 ohm^{-1} cm^2 mol^{-1}$

Answer 1

Answer: (C)

Electrolyte conductivity, $k = \frac{1}{R}\times $ cell constant
$ = \frac{1}{100} \times 0.037 = 3.7\times 10^{-4} ohm^{-1} cm^{-1}$
$\therefore $ Molar conductivity $= \frac {1000 \times 3.7\times 10^{-4}}{0.05}$
$ = 7.4\,ohm^{-1} cm^2\,mol^{-1}$.