`K_(f) = 1.82 K kg mol^(-1)`, 80 g of solute dissolved in 400 g water, molecular weight of solute is 62.Calculate freezing point of solution.

Options

A. 268.31K
B. 266.31K
C. 267.28K
D. 270.31K

Answer

C

Explanation

`Delta T_(f) =K_(f) m = K_(f) xx ( W_(2))/( M_(2)) xx(1000)/( W_(1))`
`DeltaT_(f) = 1.82 xx ( 80xx 1000)/( 62 xx 40) =5.6875`
`Delta T_(f) =T_(0) - T_(s)`
`:. T_(s) =T_(0) -Delta T_(f) = 273 - 5.6875 = 261.3125K `