The elevation in boling point of an aqueous solutions of NaCl is 0.01 `.^(@) C` . If its van't Hoff factor is 1.92 ,the molality of NaCl solution is
( `K_(b)` for water = 0 . 52 k kg `mol^(-1)`)

Options

A. 0.01m
B. 0.001m
C. 0.005m
D. 0.02m

Answer

A

Explanation

Given , `Delta T_(b) = 0 . 001.^(@) C = 0 . 01 K `
i = 1.92
`K_(b) =0 .52 K kg mol^(-1)`
As we know that,
`Delta T_(b) = iK_(b) xx m `
`:. m = (0.01)/(1 . 92 xx 0. 52) mol//kg`
m (molality) = 0 . 01 mol / kg or 0 . 01 m