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

`(Delta p)/(p) = (n_(B))/(n_(B)+n_(A))`
`0.0125 = (n_(B))/(n_(B)+n_(A))`
`(125)/(10000)=(n_(B))/(n_(B)+n_(A))` or `(1)/(80)=(n_(B))/(n_(A)+n_(B))`
Let `n_(B)=1 therefore n_(A) = 79`
Mass of water `= 79xx18=1422 g = 1.422 kg`
`therefore` Molaity of solution `= (1mol)/(1.422 kg)=0.70 m`