Question

The resistivity of a $0.8 M$ solution of an electrolyte is $5 \times 10^{-3} \Omega cm$ Its molar conductivity is ____ $\times 10^4\, \Omega^{-1}\, cm ^2 \,mol ^{-1}$

Hint:

The molar conductivity is calculated using the relationship between resistivity and molarity. Keep in mind the dimensions of the quantities involved.

Answer 1

Correct Answer: 25

${\Lambda }_m=\frac{\kappa \times 1000}{M}$
${\Lambda }_m=\frac{1}{\rho }\times \frac{1000}{M}$
$\frac{1}{5\times 10^{-3}}\times \frac{1000}{0.8}$
So , correct answer is $25\times 10^4{\Omega }^{-1}c{m}^{-2}mo{l}^{-1}$

Answer 2

The molar conductivity \( \Lambda_m \) is given by the formula: \[ \Lambda_m = \frac{k \times 1000}{M} \] where \( k \) is the resistivity and \( M \) is the molarity of the solution. Also, we have the formula: \[ \Lambda_m = \frac{1000}{\rho} \quad \text{where} \quad \rho = \text{resistivity} \] Now, using the given values: \[ \Lambda_m = \frac{1}{\rho} \times 1000 = \frac{1}{5 \times 10^{-3}} \times 1000 \] Substitute the value of molarity \( M = 0.8 \): \[ \Lambda_m = \frac{1000}{5 \times 10^{-3}} \times 0.8 = 25 \times 10^4 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \] Thus, the molar conductivity is \( 25 \times 10^4 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \).