The elevation in boiling point of a solution of `13.44`g of `CuCl_(2)`(molecular weight =`134.4,k_(b)=0.52K "molality"^(-1))` in `1` kg water using the following information will be:

Options

A. 0.16
B. 0.05
C. 0.1
D. 0.2

Answer

A

Explanation

(i) `i = ("No. of particle after ionisation")/("No. of particle before ionisation")`
(ii) `DeltaT_b = ixxK_bxxm`
`CuCl_2 to Cu^(2+) + 2Cl^(-)`
`{:(1,0,0),((1-alpha),alpha,2alpha):}`
`i=(1 + 2alpha)/1 , i=1 + 2alpha`
Assuming 100% ionization
So, i = 1 + 2 = 3
`DeltaT_b = 3 xx 0.52 xx 0.1 = 0.156 ~~0.16 [m=(13.44)/134.4=0.1]`