A solution of 0.5 g of a solute ( molar mass =150`gmol^(-1)`) in 50 g of a solvent yields a boiling point eleation of 0.40 K. Another solution of 0.60 g of an unknown solute in the same mass of solvent exhibits a boiling point elevation of 0.8 K. The molar mass of unknown solute is

Options

A. `60gmol^(-1)`
B. `90gmol^(-1)`
C. `120gmol^(-1)`
D. `180gmol^(-1)`

Answer

B

Explanation

`(DeltaT_(b_(1)))/(Delta_(b_(2)))prop (m_(1))/(m_(2))prop(("moles")_(1))/(("moles")_(2))[( :. "mass of solvent is same"]`
`(0.40)/(0.80)=(0.5xx x)/(150xx0.60)`
`=0.4x=36`
`x=90`