`H_(2)O (1) hArr H_(2) O(g)` at `373K,DeltaH^(@)=8.31kcal mol^(-1)` Thus, boiling point of 0.1 molal sucrose solution is

Options

A. `373.52K`
B. `373.052K`
C. `373.06K`
D. `374.52K`

Answer

C

Explanation

`H_(2)O(l)` changes to steam `H_(2)O (g)`at`373K`.Thus , this represent latent heat of vaporization .`K_(b)`(molal elevaion constant)is related to `DeltaH^(@)` and boiling point by equation,
`K_(b)=(RT_(0)^(2))/(1000cancelOH^(@)`
Here,`DeltaH^(@)` is in energy unit per gram of solvent.
`=(8.31)/(18)kcal^(-1)`
`:.K_(b)=(0.002xx(373)^(2))/(1000xx((8.31)/(18)))`
`=(278.25)/(461.66)=0.60^(0)mol^(-1)kg`
`DeltaT`(surrose solution = molarity `xx K_(b)=0.1xx0.60=0.06^(0)`
`:.` Boiling point of solution `= 373+0.06^(@)=373.06K`