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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 hg
The weight of water=850 g
     
 The weight of Urea=50 g 

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

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   $$n_2$$=$$\cfrac{50}{60}$$=0.83

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
      $$\cfrac{P_1^0-P_1}{P_1^0}=\cfrac{n_2}{n_1+n_2}$$

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 Hg

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

Relative lowering=$$0.0173$$ 

Chemistry

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