1.

The relative lowering of vapour pressure of an aqueous solution containing non-volatile solute is 0.0125. The molality of the solution is

Answer»

0.7
0.5
0.6
0.8

Solution :Relative lowering of VAPOUR pressure
= mole fraction of the SOLUTE in the solution,
i.e., `(n_(2))/(n_(1)+n_(2))=0.0125`
Molality `=n_(2)" when "n_(1)=1000/18 " mole"`
= 55.55 mole
`(n_(1)+n_(2))/(n_(2))=(1)/(0.0125)=80or(n_(1))/(n_(2))+1=80`
`"or"(n_(1))/(n_(2))=79 or n_(2)=(n_(1))/(79)=(55.55)/(79)=0.70`


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