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. `0.70`
B. `0.50`
C. `0.60`
D. `0.80`

Answer

A

Explanation

Relative lowering of vapour pressure of an aqueous solution containing nonvolatile solute is equal to mole fraction of solute.
`therefore (P^(@)-P_(s))/(P^(@))=(n)/(n+N)=0.0125`
`rArr (n+N)/(n)=(1)/(0.0125)rArr (N)/(n)=(1)/(0.0125)-1=(0.9875)/(0.0125)`
`therefore (N)/(n)=(0.9875)/(0.0125)`
Now, molality `=(0.0125xx1000)/(0.9875xx18)=0.70`.