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

Answers (1)

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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