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 m`
B. `0.590 m`
C. `0.80 m`
D. `0.40 m`

Answer

a

Explanation

Relative lowering of vapour pressure of an acqueous solution containing non-volatile solute a = is mole fraction of solute
As `(p_(0) - p_(s))/(p_(0)) = (n)/(n + N) = 0.0125`
`rArr (n + N)/(N) = (1)/(0.0125) - (N)/(n) = (1-1)/(0.0125) = (0.9875)/(0.0125)`
So, `(N)/(n) = (0.9875)/(0.0125)`
Now, molality `= (0.0125xx1000)/(0.9875xx18) = 0.70`