Vapour pressure of pure water at $298 \, K$ is $23.8 \, mm \, Hg$. $50 \, g$ of urea $(NH_2CONH_2)$ is dissolved in $850 \, g$ of water. Calculate the vapour pressure of water for this solution and its relative lowering.
(N/A) Given: Vapour pressure of pure water,$p_1^0 = 23.8 \, mm \, Hg$.
Weight of water,$w_1 = 850 \, g$.
Weight of urea,$w_2 = 50 \, g$.
Molar mass of water,$M_1 = 18 \, g \, mol^{-1}$.
Molar mass of urea,$M_2 = 60 \, g \, mol^{-1}$.
Number of moles of water,$n_1 = \frac{850}{18} = 47.22 \, mol$.
Number of moles of urea,$n_2 = \frac{50}{60} = 0.833 \, mol$.
Relative lowering of vapour pressure is given by $\frac{p_1^0 - p_1}{p_1^0} = \frac{n_2}{n_1 + n_2}$.
$\frac{p_1^0 - p_1}{p_1^0} = \frac{0.833}{47.22 + 0.833} = \frac{0.833}{48.053} = 0.0173$.
Now,$23.8 - p_1 = 0.0173 \times 23.8 = 0.4117$.
$p_1 = 23.8 - 0.4117 = 23.388 \, mm \, Hg \approx 23.4 \, mm \, Hg$.
Thus,the vapour pressure of the solution is $23.4 \, mm \, Hg$ and the relative lowering is $0.0173$.
Weight of water,$w_1 = 850 \, g$.
Weight of urea,$w_2 = 50 \, g$.
Molar mass of water,$M_1 = 18 \, g \, mol^{-1}$.
Molar mass of urea,$M_2 = 60 \, g \, mol^{-1}$.
Number of moles of water,$n_1 = \frac{850}{18} = 47.22 \, mol$.
Number of moles of urea,$n_2 = \frac{50}{60} = 0.833 \, mol$.
Relative lowering of vapour pressure is given by $\frac{p_1^0 - p_1}{p_1^0} = \frac{n_2}{n_1 + n_2}$.
$\frac{p_1^0 - p_1}{p_1^0} = \frac{0.833}{47.22 + 0.833} = \frac{0.833}{48.053} = 0.0173$.
Now,$23.8 - p_1 = 0.0173 \times 23.8 = 0.4117$.
$p_1 = 23.8 - 0.4117 = 23.388 \, mm \, Hg \approx 23.4 \, mm \, Hg$.
Thus,the vapour pressure of the solution is $23.4 \, mm \, Hg$ and the relative lowering is $0.0173$.
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