(a) A 5% solution (by mass) of cane - sugar in water has freezing point of 271 K. Calculate the frttzing point of 5% solution (by mass) of glucose in water if the freezing point of pure water is 273.15 K. [Molecular masses : Glucose `C_(6)H_(12)O_(6)` : 180 amu, Cane - sugar `C_(12)H_(12)O_(11) :` 342 amu]
(b) State Henry's law and mention two of its important applications.

Explanation

5% by mass cane-sugar = 5 g cane-sugar (mw = 342 g/litre)
`n_(B) = (5)/(342) ` and `w_(A) = 95 g`
`DeltaT_(f)= 273.15-271= 2.15 K`
`K_(f)= (DeltaT_(f)w_(A))/(n_(B)xx1000)=(2.15xx95)/(5/(342)xx1000)`
= 13.97 K kg `mol^(-1)`
5% by mass glucose = 5 g glucose (mw = 18/mol) in 100 g solution
`DeltaT_(f)=K_(f)xx(n_(B)1000)/(w_(A))=13.97xx(5)/(180)xx(1000)/(95)`
=4.085 K
`T_(f)= 273.15-4.085`
Freezing point of glucose solution
(b) According to Henry.s law, the partial pressure of a gas in vapour phase (p) is directly proportional to the mole fraction (x) of the gas in the solution. Henry.s law is applied in the production of carbonated beverages and for the deep-sea-divers.