The molal boiling point constant for water is`0.513^(@)C kgmol^(-1)`. When `0.1`mole of sugar is dissolved in `200`ml of water , the solution boils under a pressure of one atmosphere at

Options

A. `100.513^(@)C`
B. `100.0513^(@)C`
C. `100.256^(@)C`
D. `101.025^(@)C `

Answer

C

Explanation

`DeltaT_(b)= ( n_(B))/(W_(A)) xx K_(b) xx 1000`
`(0.1 xx 0.513)/( 200) xx 1000 =0.2565^(@)C `
`Delta T _(b)` = Boiling Pt. of solution - Boiling pt.of solvent
`:. ` B.P. Of solution =B.P. of solvent `+ DeltaT_(b)`
`= 100 + 0.2565 = 100.2565^(@)C `