Calculate the freezing point of a solution containing 60 g glucose (Molar mass = 180 g `mol^(-1)`) in 250 g of water . (`K_(f)` of water = `1.86 K kg mol^(-1)`)

Options

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

Answer

B

Explanation

Freezing point of solvent = Molal
depression constant ` xx ("Mass solute")/("Molar mass ")xx(1000)/("Mass of solvent")`
`DeltaT_(f) = K_(f) xx (W_(B))/(M_(B)) xx (1000)/(W_(A))`
` = 1.86xx (60)/(180) xx (1000)/(250)`
=2.48K
Freezing point = 273.15 K -2.481
=270.67K.