The resistance of conductivity cell with cell constant 1.14 cm$$^{-1}$$, containing 0.001M KCl at 298 K is 1500$$\Omega$$. The molar conductivity of 0.001M KCl solution at 298 K in S cm$$^2$$ mol$$^{-1}$$ is _________. (Integer answer)

We know that the specific conductivity (also called conductance per centimetre, denoted by $$\kappa$$) of a solution is obtained from the resistance $$R$$ of the conductivity cell by the relation

$$\kappa \;=\; \text{Cell constant}\;\times\; \dfrac{1}{R}$$

Here the cell constant is given as $$1.14\;\text{cm}^{-1}$$ and the measured resistance is $$R = 1500\;\Omega$$. Substituting these values, we get

$$\kappa \;=\; 1.14\;\text{cm}^{-1}\;\times\;\dfrac{1}{1500\;\Omega}$$

First calculate the reciprocal of the resistance:

$$\dfrac{1}{1500\;\Omega} = 0.000\,666\,7\;\text{S}$$

(Since $$1/\Omega$$ is the unit Siemens, S.) Multiplying by the cell constant:

$$\kappa = 1.14 \times 0.000\,666\,7\;\text{S cm}^{-1}$$

Carrying out the multiplication,

$$\kappa = 0.000\,760\;\text{S cm}^{-1}$$

Now we need the molar conductivity $$\Lambda_m$$. The defining formula connecting molar conductivity with specific conductivity is

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

where

• $$\kappa$$ is in $$\text{S cm}^{-1}$$,
• 1000 converts litres to cubic centimetres because $$1\;\text{L} = 1000\;\text{cm}^3$$,
• $$C$$ is the molarity in $$\text{mol L}^{-1}$$.

The concentration of the KCl solution is $$C = 0.001\;\text{M} = 0.001\;\text{mol L}^{-1}$$. Substituting all the known values,

$$\Lambda_m = \dfrac{0.000\,760\;\text{S cm}^{-1}\;\times\;1000}{0.001\;\text{mol L}^{-1}}$$

First multiply $$\kappa$$ by 1000:

$$0.000\,760 \times 1000 = 0.760\;\text{S cm}^2$$

Next divide by the concentration:

$$\Lambda_m = \dfrac{0.760\;\text{S cm}^2}{0.001} = 760\;\text{S cm}^2\;\text{mol}^{-1}$$

Thus the molar conductivity of the $$0.001\;\text{M}$$ KCl solution at 298 K is

$$\boxed{760\;\text{S cm}^2\;\text{mol}^{-1}}$$

So, the answer is $$760$$.