Question

If 0.01 M solution of an electrolyte has a resistance of 40 Ohm's in a cell having a cell constant of $0.4cm^{- 1},$ then its molar conductance in $Ohm^{- 1}cm^{2}mol^{- 1}$ is

• A. 10
• B. $10^{2}$
• C. $10^{3}$
• D. $10^{4}$

Answer 1

Answer: (C)

$10^{3}$

Answer 2

$\kappa = \frac{1}{R} \times \frac{l}{a}$

(where $\kappa =$specific conductance, R = Resistance, $\frac{l}{a} =$cell constant)

$\mathbf{\therefore}\mathbf{\kappa =}\frac{\mathbf{1}}{\mathbf{40}}\mathbf{\times}\mathbf{0.4 = 0.01o}\mathbf{h}\mathbf{m}^{\mathbf{-}\mathbf{1}}\mathbf{c}\mathbf{m}^{\mathbf{-}\mathbf{1}}$

Strength of the solution = 0.01M

$\therefore$Volume of solution containing one mole of the electrolyte = 1,00,000 cc.

$\therefore$Molar conductance $= \kappa \times V$in cc.$= 0.01 \times 1,00,000 = 10^{3}Ohm^{- 1}cm^{2}mol^{- 1}$