In certain hill station, the boiling point of pure water is found to be 94.8°C, then the amount of NaCl must be added to 1kg of water to bring boiling point back to 100°C is (K_b of water = 0.52 kg mol^-1)

Let amount of NaCl = x

Molar mass of NaCl = 58.5 g

i = 2 

Wt. of solvent = 1 kg

K_b of water = 0.52 kg mol^-1

${\mathrm{\Delta T}}_{\mathrm{b}}=100-94.8=5.2$

${\mathrm{\Delta T}}_{\mathrm{b}}=\mathrm{i}\times {\mathrm{k}}_{\mathrm{b}}\times \mathrm{m}$

$\mathrm{m}=\frac{{\mathrm{\Delta T}}_{\mathrm{b}}}{\mathrm{i}\times {\mathrm{k}}_{\mathrm{b}}}=\frac{5.2}{2\times 0.52}=5 \mathrm{molal}$

Molality = $\frac{\mathrm{no}. \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{wt}. \mathrm{of} \mathrm{solvent}}$

$5=\frac{{\displaystyle \frac{x}{58.5}}}{1}$

$\mathrm{x}=5\times 58.5=292.5 \mathrm{g}$

Therefore, the correct answer is B) 292.5 g.