1.

An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molecular mass of the solute?

Answer»

Solution :Vapour pressure of PURE water at the boiling point `(p^(@))=" 1 atm= 1.013 bar"`
Vapour pressure of solution `(p_(s))="1.004 bar , MASS of solute "(w_(2))=2 g`
`"Mass of solution = 100 g,Mass of solvent = 98 g"`
APPLYING Raoult's law for dilute solution `"(being %)"=(p^(@)-p_(s))/(p^(@))-(n_(2))/(n_(1))~=(n_(2))/(n_(1))=(w_(2)//M_(2))/(w_(1)//M_(1))=(w_(2))/(M_(2))xx(M_(1))/(w_(1))`
`THEREFORE((1.013-1.004))/("1.013 bar")=(2g)/(M_(2))xx("18 g mol"^(-1))/("98 g")"or"M_(2)=(2xx18)/(98)xx(1.013)/(0.009)" g mol"^(-1)="41.35 g mol"^(-1)`


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