JEE MAIN - Chemistry (2023 - 6th April Morning Shift - No. 18)
Mass of Urea $$\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)$$ required to be dissolved in $$1000 \mathrm{~g}$$ of water in order to reduce the vapour pressure of water by $$25 \%$$ is _________ g. (Nearest integer)
Given: Molar mass of N, C, O and H are $$14,12,16$$ and $$1 \mathrm{~g} \mathrm{~mol}^{-1}$$ respectively
Answer
1111
Explanation
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 $$1111 \, \text{g}$$.
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