JEE Main 2023 — Solutions Question with Solution

Given:

• Vapor pressure reduction: $25\%$ ($$0.75$$ times the vapor pressure of pure water)


• Molar mass of water ($${\text{H}}_2\text{O}$$): $$18\text{g/mol}$$


• Mass of solvent (water): $$1000\text{g}$$

Using Raoult's law:

$$\frac{P^0-P_s}{P_s}=\frac{n_{\text{solute}}}{n_{\text{solvent}}}=\frac{\frac{x}{M_{\text{urea}}}}{\frac{1000}{M_{\text{water}}}}=\frac{P^0-0.75P^0}{0.75P^0}$$

Solving for (x):

$$\frac{x}{60}=\frac{10000}{9}$$ $$x=1111.11111\approx 1111\text{g}$$

So, the mass of urea required to be dissolved in $$1000\text{g}$$ of water is $$1111\text{g}$$.