A solution of sucrose (molar mass `= 342 g mol^(-1))` has been prepared by dissolving 68.5g of sucrose in 1000g of water. The freezing point of the solution obtained will be: `(K_(f)` for water `= 1.86 K kg mol^(-1))`

Options

A. 0.72
B. 0.8
C. 0.92
D. 0.88

Answer

D

Explanation

Molality of solution = 0 .2 m
`Delta T_(f)` (calculated) ` = 1.86 xx 0.2 = 0.372`
`Delta T_(f)` (observed ) = 0.698
`therefore" " I = (0.698)/(0.372) = 1.88`
`Na to Na^(+) + Cl^(-)`
It `alpha` is the degree of dissociaiton.
` 1- alpha " " alpha`
Total moles ` = 1 - alpha + alpha + alpha = 1 + alpha`
`i = (1 + alpha)/(1) = 1.88`
`therefore " "a = 0.88 or 88%`