1.

Vapour pressure of water at 293 K is 17.51 mm. Lowering of vapour pressure of a sugar solution is 0.0614 mm. Calculate (i) Relative lowering of vapour pressure (ii) Vapour pressure of the solution and (iii) Mole fraction of water .

Answer»

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Solution :Here, we are GIVEN that Vapour pressure of water `(p^(@))=17.51mm`
Lowering of vapour pressure `(p^(@)-p_(S))=0.0614mm`
(i)`"Relative lowering of vapour pressure "=(p^(@)-p_(s))/(p^(@))=("0.0614 mm")/("17.51 mm")="0.00351"`
(ii) Vapour pressure of the solution `(p_(s))`
`p^(@)-p_(s)="0.0614 mm,"p^(@)="17.51 m"`
`THEREFORE"17.51mm "-p_(s)="0.0614 mmor"p_(s)="17.51 mm "-0.0614 mm = 17.4468 mm"`
(III) To calculate mole fraction of water :
By Raoult's law `(p^(@)-p_(s))/(p^(@))=(n_(2))/(n_(1)+n_(2))=x_(2),` mole fraction of solution, i.e., `x_(2)=(p^(@)-p_(s))/(p^(@))=0.00351`
`therefore"Mole fraction of solvent (water )," x_(1)=1-x_(2)=1-0.00351=0.99649`


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