Resistance of a conductivity cell (cell constant 129 m$$^{-1}$$) filled with 74.5 ppm solution of KCl is 100 $$\Omega$$ (labelled as solution 1). When the same cell is filled with KCl 149 ppm solution of KCl, the resistance is 50 $$\Omega$$ (labelled as solution 2). The ratio of molar conductivity of solution 1 and solution 2 is i.e. $$\frac{\Lambda_1}{\Lambda_2} = x \times 10^{-3}$$. The value of x is _____ (Given, molar mass of KCl is 74.5 g mol$$^{-1}$$)

We are given a conductivity cell with cell constant $$G^* = 129$$ m$$^{-1}$$ and two KCl solutions. We need to find the ratio of their molar conductivities.

For Solution 1: concentration = 74.5 ppm, resistance = 100 $$\Omega$$.

For Solution 2: concentration = 149 ppm, resistance = 50 $$\Omega$$.

First, we find the conductivity ($$\kappa$$) of each solution using $$\kappa = \frac{G^*}{R}$$:

$$\kappa_1 = \frac{129}{100} = 1.29 \text{ S m}^{-1}$$

$$\kappa_2 = \frac{129}{50} = 2.58 \text{ S m}^{-1}$$

Now we find the molar concentrations. Since ppm means mg per litre (for dilute aqueous solutions) and the molar mass of KCl is 74.5 g/mol:

$$c_1 = \frac{74.5 \text{ mg/L}}{74.5 \text{ g/mol}} = \frac{74.5 \times 10^{-3}}{74.5} = 10^{-3} \text{ mol/L} = 1 \text{ mol/m}^3$$

$$c_2 = \frac{149 \text{ mg/L}}{74.5 \text{ g/mol}} = \frac{149 \times 10^{-3}}{74.5} = 2 \times 10^{-3} \text{ mol/L} = 2 \text{ mol/m}^3$$

The molar conductivity is given by $$\Lambda_m = \frac{\kappa}{c}$$ (where $$c$$ is in mol/m$$^3$$ when $$\kappa$$ is in S/m):

$$\Lambda_1 = \frac{1.29}{1} = 1.29 \text{ S m}^2 \text{ mol}^{-1}$$

$$\Lambda_2 = \frac{2.58}{2} = 1.29 \text{ S m}^2 \text{ mol}^{-1}$$

The ratio is:

$$\frac{\Lambda_1}{\Lambda_2} = \frac{1.29}{1.29} = 1$$

We are told $$\frac{\Lambda_1}{\Lambda_2} = x \times 10^{-3}$$. Therefore $$x \times 10^{-3} = 1$$, which gives $$x = 1000$$.

Hence, the correct answer is 1000.