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