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

Options

A. 0.7
B. 0.5
C. 0.6
D. 0.8

Answer

A

Explanation

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`