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Vapour pressure in $\mathrm{mm} \mathrm{Hg}$ of 0.1 mole of urea in $180 \mathrm{~g}$ of water at $25^{\circ} \mathrm{C}$ is (The vapour pressure of water at $25^{\circ} \mathrm{C}$ is $24 \mathrm{~mm} \mathrm{Hg}$ )
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The correct answer is:
$23.76$
From Raoult's law for very dilute solution.
$$
\begin{aligned}
\frac{p^{\circ}-p_s}{p^{\circ}} & =\frac{w}{m} \times \frac{M}{W} \\
\frac{24-p_s}{24} & =0.1 \times \frac{18}{180} \\
24-p_s & =0.24 \\
\therefore \quad p_s= & 24-0.24=23.76 \mathrm{~mm} \mathrm{Hg}
\end{aligned}
$$
$$
\begin{aligned}
\frac{p^{\circ}-p_s}{p^{\circ}} & =\frac{w}{m} \times \frac{M}{W} \\
\frac{24-p_s}{24} & =0.1 \times \frac{18}{180} \\
24-p_s & =0.24 \\
\therefore \quad p_s= & 24-0.24=23.76 \mathrm{~mm} \mathrm{Hg}
\end{aligned}
$$
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