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# 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.

2.9    Vapour pressure of pure water at $298 \; K$ is $\inline 23.8\; mm \; Hg.$ $\inline 50\; g$ of urea $\inline (NH_{2}CONH_{2})$ is dissolved in $\inline 850 \; g$ of water. Calculate the vapour pressure of water for this solution and its relative lowering.

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Given that vapour pressure of pure water,  $p^{\circ}_w = 23.8\ mm\ of\ Hg$

Moles of water  :

$= \frac{850}{18} = 47.22$

Moles of urea  :

$= \frac{50}{60} = 0.83$

Let the vapour pressure of water be pw.

By Raoult's law, we get :

$\frac{p^{\circ}_w - p_w}{p^{\circ}_w} = \frac{n_2}{n_1\ + n_2}$

or                         $\frac{23.8 - p_w}{23.8} = \frac{0.83}{47.22\ + 0.83}=0.0173$

or                          pw  =   23.4 mm of Hg.

Relative lowering :-  Hence, the vapour pressure(v.P) of water in the solution = 23.4 mm of Hg

and its relative lowering = 0.0173.

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