A solution prepared by dissolving a `2.50g` sample of a unknown compound dissolved in `34.0g` of benzene `C_(6)H_(6)` boils at `1.38^(@)C` higher than pure benzene. Which expression gives the molar mass of the unknown compound?
`{:("Compound",K_(b)),(C_(6)H_(6),2.53^(@)Cm^(-1)):}`

Options

A. `2.53 xx (2.50)/(1.38)`
B. `1.38 xx (34.0)/(2.53)xx2.50`
C. `2.50xx10^(3)xx (2.53)/(34.0)xx(1)/(1.38)`
D. `2.50 xx 10^(3) xx (1.38)/(34.0)xx2.53`

Answer

C

Explanation

`1.38 = (2.5)/(34xxM) xx 10^(3) xx 2.53`
`M = (2.5 xx 10^(3) xx 2.53)/(34 xx 1.38)`