Two 5 molal solutions are prepared by dissolving a non-electroylete, non -volatile solute separately in the solvents X and Y. The molecular weights of the solvents are `M_(x)` and `M_(Y)`, respectively where `M_(X)=4/3M_(Y)`. The relative lowering of vapour pressure of the solution in X is "m" times that of the solution in Y. Given that the number of moles of solute is very small in comparision to that of solvent, the value of "m" is :

Options

A. `3/4`
B. `4/3`
C. `1/2`
D. `1/4`

Explanation

Molality=`("Number of moles of solute")/("Mass of solvent (in kg)")`
Number of moles of `=5xx1=5` (in both in solvents)
Number of moles of solvent `x=1000/(M_(x))`
Number of moles of solvent `Y=1000/(M_(Y))`
Relative lowering in vapour pressure is given as `(p^(@)-p_(s))/(p_(s))=(n_(2))/(n_(1))`
`((p^(@)-p_(s))/(p_(s)))_("solution in X")=5/(1000/M)=(5M_(x))/1000`
`((p^(@)-p_(s))/(p_(s)))_("solution in Y")=5/(1000/(M_(Y)))=(5M_(x))/1000`
According to question `1/mxx(5M)/1000=(5M_(Y))/1000`
`1/mxx5xx3/4M_(Y)=M_(Y)` (Given `M_(x)=3/4M_(Y)" ":.m=3/4`