At 298 K, the molar conductivity of an $$x\%$$ (w/w) MX solution is 123.5 S cm$$^{2} mol^{-1}$$. The conductance of same solution is $$1.9 \times 10^{-3} S$$. The value of x is ______ $$\times 10^{-2}$$.
(Given: Cell constant = 1.3 cm$$^{-1}$$, molar mass of MX = 75 $$g mol^{-1}$$, density of aqueous solution of $$MX$$ at 298 K is 1.0 $$g mL^{-1}$$)

To determine the percentage by weight of the solution, we use the relationships between conductance, conductivity, and molar conductivity.

The conductivity $$\kappa$$ is related to the conductance $$G$$ and the cell constant $$G^*$$ by

$$\kappa=G\times G^*.$$

Substituting the given values,

$$\kappa=(1.9\times10^{-3},\text{S})\times(1.3,\text{cm}^{-1})=2.47\times10^{-3},\text{S cm}^{-1}.$$

The molar conductivity is given by

$$\Lambda_m=\frac{1000\kappa}{M}.$$

Using $$\Lambda_m=123.5,\text{S cm}^2\text{ mol}^{-1}$$ and $$\kappa=2.47\times10^{-3},\text{S cm}^{-1}$$,

$$123.5=\frac{1000\times2.47\times10^{-3}}{M}.$$

Hence,

$$M=\frac{2.47}{123.5}=0.02,\text{mol L}^{-1}.$$

The relation between molarity, density, and percentage by weight is

$$M=\frac{x\times d\times10}{\text{Molar Mass}},$$

where $$x$$ is the percentage by weight.

Substituting the given values $$M=0.02,\text{mol L}^{-1}$$, $$d=1.0,\text{g mL}^{-1}$$, and molar mass $$=75,\text{g mol}^{-1}$$,

$$0.02=\frac{x\times1.0\times10}{75}.$$

Therefore,

$$x=\frac{0.02\times75}{10}=\frac{1.5}{10}=0.15.$$

Expressing this in the form $$n\times10^{-2}$$,

$$0.15=15\times10^{-2}.$$

Hence, the value of $$x$$ is $$15$$.