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^(-1)`
C. `17.3 g mol^(-1)`
D. `20.4 g mol^(-1)`

Answer

C

Explanation

First boiling point of water `= 100^(@)C`
Final boiling point of water `= 100.52^(@)`
`w=3g, W=200g, K_(b)=0.6 K-kg mol^(-1)`
`Delta T_(b)=100.52-100=0.52^(@)C`
`m=(K_(b)xx wxx1000)/(Delta T_(b)xx W)`
`=(0.6xx3xx1000)/(0.52xx200)=(1800)/(104)=17.3 gmol^(-1)`.