Boiling point of water at $750\text{ mm Hg}$ pressure is $99.68^\circ\text{C}$. How much sucrose ($\text{molar mass} = 342\text{ g mol}^{-1}$) is to be added to $500\text{ g}$ of water such that it boils at $100^\circ\text{C}$? ($K_b\text{ for water} = 0.52\text{ K kg mol}^{-1}$)

$105.2\text{ g}$

The correct answer is Option (3) → $105.2\text{ g}$ ##

$\Delta T_b = K_b \times m$

$0.32 = 0.52 \times \text{W} \times \frac{1000}{342 \times 500}$

$\text{W} = 0.32 \times 342 \times \frac{500}{0.52 \times 1000}$

$= 105.23\text{ g}$

$\text{Weight of sucrose to be added} = 105.23\text{ g}$