' W ' g of a non-volatile electrolyte solid solute of molar mass '...

Given the vapour pressure of pure water $P^o = 640$ mm Hg and vapour pressure of solution $P_s = 600$ mm Hg.
According to Raoult's law for an electrolyte solution, the relative lowering of vapour pressure is given by $\frac{P^o - P_s}{P^o} = i \cdot x_2$, where $i$ is the van't Hoff factor and $x_2$ is the mole fraction of the solute.
Substituting the values: $\frac{640 - 600}{640} = i \cdot x_2$
$\frac{40}{640} = i \cdot x_2 \implies i \cdot x_2 = \frac{1}{16}$ ... (1)
The elevation in boiling point is given by $\Delta T_b = i \cdot K_b \cdot m$, where $m$ is the molality.
Given $T_b = 375$ K and $T_b^o = 373$ K, so $\Delta T_b = 375 - 373 = 2$ K.
Molality $m = \frac{W}{M \times 100} \times 1000 = \frac{10W}{M}$ (since mass of water = $100$ mL $\times 1$ g/mL = $100$ g).
Substituting in the elevation formula: $2 = i \cdot 0.52 \cdot \frac{10W}{M}$
$i = \frac{2M}{5.2W} = \frac{20M}{52W} = \frac{5M}{13W}$ ... (2)
Substitute the value of $i$ from (2) into (1):
$\frac{5M}{13W} \cdot x_2 = \frac{1}{16}$
$x_2 = \frac{1}{16} \cdot \frac{13W}{5M} = \frac{13}{80} \cdot \frac{W}{M}$
This can be rewritten as $x_2 = \frac{1.3}{8} \times \frac{W}{M}$.