  Question

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.

Solution

Vapour pressure of water,$$P_1=23.8$$ m of hgThe weight of water=850 g      The weight of Urea=50 g The molecular weight of water(H_2O)=$$1\times 2+16=18\ g\ mol^{-1}$$molecular weight of urea $$(NH_2CONH_2)=2N+4H+C+O=2\times 14+4\times 1+12+16=60\ g\ mol^{-1}$$Use formula          number of moles=$$\cfrac{\text{mass}}{\text{molar mass}}$$number of moles of water   $$n_1$$=$$\cfrac{850}{18}$$=$$47.22$$number of moles of urea   $$n_2$$=$$\cfrac{50}{60}$$=0.83Now, 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      $$\cfrac{P_1^0-P_1}{P_1^0}=\cfrac{n_2}{n_1+n_2}$$plug the values we get$$\cfrac{23.8-P_1}{23.8}=\cfrac{0.83}{47.22+0.83}$$$$\cfrac{23.8-P_1}{23.8}=0.0173$$$$23.8-P_1$$=$$23.8\times 0.0173$$$$P_1=23.4$$ m HgVapour Pressure of water in the given solution$$=23.4$$ mm of HgRelative lowering=$$0.0173$$ Chemistry

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