Question

The boiling point of water becomes 100.52 degree C. If 3g of a non-volatile solution is dissolved in 20ml of it. Calculate molar mass of solution ​

Answer 1

Answer:

`12.2g mol^(-1)`

`15.4 g mol^(-1)`

`17.3 g mol^(-1)`

`20.4 g mol^(-1)`

Answer :

C

Solution :

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)`.