JEE Main 2016 — Solutions Question with Solution

According to Raoult's Law

$$\frac{P^{\circ }-P_s}{P_s}=\frac{W_B\times M_A}{M_B\times W_A}...\left(i\right)$$

Here $$P^{\circ }=$$ Vapour pressure of pure solvent,

$$P_s=$$ Vapour pressure of solution

$$W_B=$$ Mass of solute,

$$W_A=$$ Mass of solvent

$$M_B=$$ Molar mass of solute,

$$M_A=$$ Molar Mass of solvent

Vapour pressure of pure water at $$100^{\circ }C$$

(by assumption $$=760$$ torr)

By substituting values in equation $$(i)$$ we get,

$$\frac{760-P_s}{P_s}=\frac{18\times 18}{180\times 178.2}...\left(ii\right)$$

On solving $$(ii)$$ we get

On solving $$(ii)$$ we get

$$P_s=752.4torr$$