45 g of ethylene glycol `(C_(2)H_(6)O_(2))` is mixed with 600 g of water. Calculate (a) the freezing point depression and (b) the freezing point of the solution.

Explanation

Depression in freezing point is related to the molality, therefore, the molality of the solution with respect to ethylene glycol.
`= ("moles of ethylene glycol")/("mass of water in kilogram")`
`therefore` Moles of ethylene glycol `= (45 g)/(62 g mol^(-1))`
= 0.73 mol
`therefore` Mass of water in `kg = (600 g)/(1000 g kg^(-1))=0.6 kg`
`therefore` Hence molality of ethylene glycol
`=(0.73 mol)/(0.60 kg)`
`= 1.2 mol kg^(-1)`
`therefore` Therefore freezing point depression,
`Delta T_(f)=1.86K kg mol^(-1)xx 1.2 mol kg^(-1)`
= 2.2 K
`therefore` Freezing point of the aqueous solution,
`= 273.15 K - 2.2 K`
= 270.95 K