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Calculate the vapour pressure of a 0.1 M urea solution at 298 K. The vapour pressure of pure water at 298 K is 20 mm hg. Assume density of solution to be 1g `mL^(-1)` |
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Answer» Correct Answer - 199.6 mm Hg 0.1 M urea solution contains 6.1 mole of urea in 1000 mL of solution. `"Mass of 1000 mL urea solution"=Vxxd=(1000mL)xx(1g mL^(-1))=1000 g` `"Mass of 0.1 mole of urea"=(0.1 mol)xx(60 g mol^(-1))=6g` `"Mass of water"(n_(A))=((994g))/((18 g mol^(-1)))=55.22 mol` `"Mole fraction os water"(x_(A)) =((55.22 mol))/((55.22 mol+0.1 mol))=0.998` `Vapour pressure of solution (P)=P_(A)^(@)X_(A)=(200mm)xx(0.998)-199.6 mm Hg`. |
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