A 5% solution (by mass) of cane sugar in water has freezing point of 27IK Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.

Explanation

We are given that
Mass of solute = 5 g , Freezing point of solution = 271 K
Freezing point of pure water = 273.15 K
We are to calculate the freezing point of 5% glucose in water.
Using the relation, `DeltaT_(f)=K_(f)xxm`
Here, molality of sugar solution, (m) = `(w_(B)xx1000)/(M_(B)xxw_(A))=(5xx1000)/(342xx100)=0.146`
`DeltaT_(f)` for sugar solution = 273.15 - 271 = 2.15 K
`thereforeK_(f)=(DeltaT_(f))/(m)=2.15/(0.146)`
Molality of glucose solution = `5/(180)xx1000/(100)=0.278`
`thereforeDeltaT_(f)` for glucose solution = `2.15/(0.146)xx0.278=4.09`
Freezing point of gflucose solution = 273.15 - 4.09 = 269.06 K