Question

The resistance of $\frac{1}{10}$ M solution is $2.5 \times 10^3 \text{ohm}$. What is the molar conductivity of solution? (cell constant = $1.25 \text{cm}^{-1}$)

Answer 1

5

Answer 2

The conductivity ($\kappa$) of the solution is related to the cell constant ($G^*$) and resistance ($R$) by the formula:

$\kappa = \frac{G^*}{R}$

Given $G^* = 1.25 \text{ cm}^{-1}$ and $R = 2.5 \times 10^3 \text{ ohm}$.

$\kappa = \frac{1.25 \text{ cm}^{-1}}{2.5 \times 10^3 \text{ ohm}} = \frac{1.25}{2500} \text{ S cm}^{-1} = 0.0005 \text{ S cm}^{-1} = 5 \times 10^{-4} \text{ S cm}^{-1}$

The molar conductivity ($\Lambda_m$) is related to the conductivity ($\kappa$) and molar concentration ($C$) by the formula:

$\Lambda_m = \frac{\kappa}{C} \times 1000$

where $\kappa$ is in S cm$^{-1}$, $C$ is in mol L$^{-1}$ (M), and $\Lambda_m$ is in S cm$^2$ mol$^{-1}$.

Given $C = \frac{1}{10} \text{ M} = 0.1 \text{ mol L}^{-1}$.

$\Lambda_m = \frac{5 \times 10^{-4} \text{ S cm}^{-1}}{0.1 \text{ mol L}^{-1}} \times 1000 \text{ cm}^3 \text{ L}^{-1}$

$\Lambda_m = \frac{5 \times 10^{-4}}{0.1} \times 1000 \text{ S cm}^2 \text{ mol}^{-1}$

$\Lambda_m = (5 \times 10^{-3}) \times 1000 \text{ S cm}^2 \text{ mol}^{-1}$

$\Lambda_m = 5 \text{ S cm}^2 \text{ mol}^{-1}$

The molar conductivity of the solution is $5 \text{ S cm}^2 \text{ mol}^{-1}$.