Vapour pressure of pure water at $298 K$ is $23.8 m m H g$ . $50 g$ of urea $(N H_{2} C O N H_{2})$ is dissolved in $850 g$ of water. Calculate the vapour pressure of water for this solution and its relative lowering.

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Solution

Vapour pressure of water, $P_{1} = 23.8$ m of hg
The weight of water=850 g

The weight of Urea=50 g

The molecular weight of water(H_2O)= $1 \times 2 + 16 = 18 g m o l^{- 1}$

molecular weight of urea $(N H_{2} C O N H_{2}) = 2 N + 4 H + C + O = 2 \times 14 + 4 \times 1 + 12 + 16 = 60 g m o l^{- 1}$

Use formula

number of moles= $\frac{mass}{molar mass}$

number of moles of water $n_{1}$ = $\frac{850}{18}$ = $47.22$

number of moles of urea $n_{2}$ = $\frac{50}{60}$ =0.83

Now, we have to calculate the vapour pressure of water in the solution. we take vapour pressure as $P_{1}$ .

Use the formula of Raoult's law
$\frac{P_{1}^{0} - P_{1}}{P_{1}^{0}} = \frac{n_{2}}{n_{1} + n_{2}}$

plug the values we get

$\frac{23.8 - P_{1}}{23.8} = \frac{0.83}{47.22 + 0.83}$

$\frac{23.8 - P_{1}}{23.8} = 0.0173$

$23.8 - P_{1}$ = $23.8 \times 0.0173$

$P_{1} = 23.4$ m Hg

Vapour Pressure of water in the given solution $= 23.4$ mm of Hg

Relative lowering= $0.0173$

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Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH2CONH2) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.

Vapour pressure of pure water at $298 K$ is $23.8 m m H g$ . $50 g$ of urea $(N H_{2} C O N H_{2})$ is dissolved in $850 g$ of water. Calculate the vapour pressure of water for this solution and its relative lowering.