The boiling point fo water `(100^(@)C` become `100.52^(@)C`, if `3` grams of a nonvolatile solute is dissolved in `200 ml` of water. The molecular weight of solute is
(`K_(b)` for water is `0.6 "K.kg mol"^(-1)`)

Options

A. `12.2g mol^(-1)`
B. 15.4 g mol
C. `17.3 g mol^(-1)`
D. 20.4 g mol

Answer

C

Explanation

First boiling point of water `=100^(@)C`
Find boiling point of water `=100.52^(@)`
`w=3g,W=200g,K_(b)=0.6K-kgmol^(-1)`
`DeltaT_(b)=100.52-100=0.52^(@)C`
`m=(K_(b)xxwxx1000)/(DeltaT_(b)xxW)`
`=(0.6xx3xx1000)/(0.52xx200)=(1800)/(104)=17.3gmol^(-1)`