The molal boiling point constant of water is `0.573^(@)C kg "mole"^(-1)`. When `0.1` mole of glucose is dissolved in `1000g` of water, the solution boils under atmospheric pressure at:

Options

A. `100.53 ^(@)C`
B. `101.06 ^(@)C`
C. `100.265^(@)C`
D. `9.47^(@)C`

Answer

C

Explanation

`Delta T = (1000 xx K xx n)/ W_(1) :. Delta T =(1000 xx 0.53 xx 2)/4000 = 0.265`
`:. T_(b) = 100 + 0.265 = 100.265 ^(@)C`