The freezing point of solution containing 60 g of glucose (Molar mass 180 g/mol) in 250 g of water (`K_(f)`-for `H_(2)O` = 1.86 K kg `mol^(-1)`) is:

Options

A. `271.67 K`
B. ` 270.67 K`
C. `274 K `
D. 270K

Explanation

Freezing point of solvent = Malal
depression constnat `xx ("Mass solute")/("Molar mass") xx (1000)/("Mass of solvent") `
`Delta T _(f) = K _(f) xx ( M _(B))/( M _(B)) xx (1000)/( W _(A) ) = 1. 86 xx (60)/(180) xx (1000)/(250) = 2. 48 K`
Freezing point `= 273.15 K - 2. 481 = 270 . 67 K`