Question

30 g of urea ( M= 60 g/mol) is dissolved in 864.0 g of water. Calculate the vapour pressure of water for this solution if vapour pressure of pure water at 298 K is 23.8 mm of Hg.

A 23.55mm

B 30.35mm

C 32.55mm

D none of these​

Answer 1

$\sf{\underline{\underline{ \purple{ \large{Answer:}}}}}$

$\sf { \blue{A. 23.55 \: mm}}$ ✔️

$\\ \\ \\ \sf{\underline{\underline{ \purple{ \large{Solution:}}}}}$

• Weight of water, w₁ = 864 g
• Weight of urea, w₂ = 30 g
• Molecular weight of water, m₁ = 18 g/mol
• Molecular weight of urea, m₂ = 60 g/mol
• Vapour pressure of pure water at 298 K is 23.8 mm of Hg.

Let vapour pressure of water be p₁.

Using Raoul's law,

$\large \bigstar {\sf { \green{ \boxed{ \gray {\sf{\dfrac{ ({p_1}^{0} - p_1) }{{{p_1}}^{0}} = \dfrac{n_2}{(n_1 - n_2)} }}}}}}$

$\\ \\ \sf \implies {\dfrac{ ({p_1}^{0} - p_1) }{{{p_1}}^{0}}} \: = \dfrac{ \frac{w_2}{m_2} }{ \frac{w_1}{m_1} - \frac{w_2}{m_2} }$

$\\ \sf \implies \dfrac{23.8 - p_1}{23.8} = \dfrac{ \frac{30}{60} }{ \frac{864}{18} - \frac{30}{60} }$

$\\ \sf \implies \dfrac{23.8 - p_1}{23.8} = \frac{ \frac{1}{2} }{ \frac{(8640 + 90)}{180} }$

$\\ \sf \implies \dfrac{23.8 - p_1}{23.8} = \frac{ \frac{1}{2} }{ \frac{8730}{180} }$

$\\ \sf \implies \dfrac{23.8 - p_1}{23.8} = \frac{ \frac{1}{2} }{ 48.5 }$

$\\ \sf \implies \dfrac{23.8 - p_1}{23.8} = \dfrac{1}{2} \times \dfrac{1}{48.5}$

$\\ \sf \implies \dfrac{23.8 - p_1}{23.8} = \dfrac{1}{97}$

$\\ \sf \implies \dfrac{23.8 - p_1}{23.8} = 0.0103$

$\\ \sf \implies 23.8 - p_1 = 0.0103 \times 23.8$

$\\ \sf \implies 23.8 - p_1 = 0.2454$

$\\ \sf \implies 23.8 = 0.2454 + p_1$

$\\ \sf \implies p_1 = 23.8 - 0.2454$

$\\ \sf \implies{ \green{ \underbrace{ \boxed {\sf{p_1 = 23.55 \: mm}}}}}$

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