The elevation in boiling point, when `13.44 g` of freshly prepared `CuCI_(2)` are added to one kilogram of water, is [Some useful data, `K_(b) (H_(2)O) = 0.52 kg K mol^(-1), "mol.wt of" CuCI_(2) = 134.4 gm]`

Options

A. `0.05`
B. `0.10`
C. `0.16`
D. `0.20`

Answer

C

Explanation

`x(CuCl_(2))=(13.44)/(134.4)=0.1`
`CuCl_(2)(aq)hArrCu^(2+)(aq)+2Cl^(-)(aq)`
`i = 3`
`DeltaT_(b)=iK_(b)xxm`
`=(3xx0.52xx0.1)/(1)=0.156~~0.16`