A 5.0 m mol dm$$^{-3}$$ aqueous solution of KCl has a conductance of 0.55 mS when measured in a cell constant 1.3 cm$$^{-1}$$. The molar conductivity of this solution is ________ mS m$$^2$$ mol$$^{-1}$$. (Round off to the Nearest Integer)

We are given a 5.0 mol m$$^{-3}$$ (i.e., 5.0 mmol dm$$^{-3}$$ = 0.005 mol dm$$^{-3}$$) aqueous solution of KCl with conductance $$G = 0.55$$ mS and cell constant $$G^* = 1.3$$ cm$$^{-1}$$.

First, we calculate the specific conductivity (conductivity): $$\kappa = G \times G^* = 0.55 \text{ mS} \times 1.3 \text{ cm}^{-1} = 0.715 \text{ mS cm}^{-1}$$.

Converting to SI units: $$\kappa = 0.715 \times 10^{-3} \text{ S cm}^{-1} = 0.715 \times 10^{-3} \times 100 \text{ S m}^{-1} = 0.0715 \text{ S m}^{-1}$$.

The molar conductivity is: $$\Lambda_m = \frac{\kappa}{c}$$, where $$c = 5.0$$ mol m$$^{-3}$$.

$$\Lambda_m = \frac{0.0715}{5.0} = 0.0143 \text{ S m}^2 \text{ mol}^{-1} = 14.3 \text{ mS m}^2 \text{ mol}^{-1}$$.

Rounding off to the nearest integer, the molar conductivity is $$14$$ mS m$$^2$$ mol$$^{-1}$$.