Question

A `5%` solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of a 5% glucose (by mass) in water. The freezing point of pure water is 273.15 K.

Answer 1

Mass of sugar in 5% (by mass) solution means 5gin 100g of solvent (water)
Molar mass of sugar`=342 " g mol"^(-1)`
Molality of sugar solution `=(5xx1000)/(342xx100)=0.146`
`therfore DeltaT_(f)` for sugar solution `=273.15-271=2.15^(@)`
`Delta T_(f)=K_(f)xxm`
`Delta T_(f)=K_(f)xx0.146 implies K_(f)=2.15//0.146`
Molality of glucose solution
`=(5)/(180)xx(1000)/(100)=0.278`
(Molar mass of glucose`=180 " g mol"^(-1))`
`Delta T_(f)=K_(f)xxm=(2.15)/(0.146)xx0.278=4.09^(@)`
`therefore` Freezing point of glucose solution
`=273.15-4.09=269.06 K`.