Ncert Exercises & Solutions

2.9 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.

It is given that vapour pressure of water, = 23.8 mm of Hg

Weight of water taken, w₁ = 850 g

Weight of urea taken, w₂ = 50 g

Molecular weight of water, M₁ = 18 g mol⁻¹

Molecular weight of urea, M₂ = 60 g mol⁻¹

Now, we have to calculate vapour pressure of water in the solution. We take vapour pressure as p₁.

Now, from Raoult’s law, we have:

$\frac{p_{1}^{0} - p_{1}}{p_{1}^{0}} = \frac{n_{2}}{n_{1} + n_{2}}$
$\Rightarrow \frac{p_{1}^{0} - p_{1}}{p_{1}^{0}} = \frac{\frac{w_{2}}{M_{2}}}{\frac{w_{1}}{M_{1}} + \frac{w_{2}}{M_{2}}}$
$\Rightarrow \frac{23.8 - p_{1}}{23.8} = \frac{\frac{50}{60}}{\frac{850}{18} + \frac{50}{60}}$
$\Rightarrow \frac{23.8 - p_{1}}{23.8} = \frac{0.83}{47.22 + 0.83}$
$\Rightarrow \frac{23.8 - p_{1}}{23.8} = 0.0173$
$\Rightarrow p_{1} = 23.4 mm of Hg$

Hence, the vapour pressure of water in the given solution is 23.4 mm of Hg and its relative lowering is 0.0173.

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