First slide

Conductivity , Molar conductivity and Equivalent Conductivity

Question

Molar conductance of 0.2 M strong electrolyte AB is $180 mho {cm}^{2} {mole}^{- 1}$ . Specific conductivity of the solution at the same temperature would be------- mho m^-1

Easy

A

$3.6 \times 10^{- 4}$
B

$3.6 \times 10^{- 3}$
C

$3.6$
D

$3.6 \times 10^{- 2}$

Solution

$\land_{m} (mho {cm}^{2} {mole}^{- 1}) = K (mho {cm}^{- 1}) \times \frac{1000}{M (mole / lit)} K = \frac{\land_{m} \times M}{1000} = \frac{180 \times 0.2}{1000} = 3.6 \times 10^{- 2} mho {cm}^{- 1}; 100 mho m^{- 1} = 1 mho {cm}^{- 1}' x' mho m^{- 1} = 3.6 \times 10^{- 2} mho {cm}^{- 1}' x' = 3.6;$

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