60 g urea is present in 222 g of its aqueous solution what is mole fraction of urea in the solution

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Volume 93, Issue 10, 15 November 2007, Pages 3392-3407

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Text Solution

Answer : 2227.6 `Nm^(-2)`

Solution : `"Moles of urea "(n_(B))=((3g))/((60" g mol"^(-1)))=0.05 mol` <br> `"Moles of urea "(n_(A))=((50g))/((18" g mol"^(-1)))=2.778 mol` <br> `"Mole fration of water "(x_(A))=(n_(A))/(n_(A)+n_(B))=((2.778 mol))/((2.778 mol+0.05 mol))=0.9823` <br> `"Vapour pressure of solution"(P)=P_(A)^(@)x_(A)` <br> `=(2267.7 Nm^(-2))xx(0.9823)xx(0.9823)2227.6 Nm^(-2).`

Text Solution

Answer : Mass of urea present =33.33 g

Solution : Given : Mole fraction of urea = 0.25 <br> Mass of urea in 200 g solution = ? <br> Mole fraction of water = 1- 0.25 =0.75 <br> Mass of urea = 0.25 `xx 60 =15 g` <br> Mass of water = 0.75 `xx 100 = 75 g` <br> `:.` Mass of solution = 15+ 75 = 90 g <br> `:.` 90 g solution contains 15 g urea <br> `:.200` g solution will contain `(200 xx 15)/(90) ` = 33.33 g urea