Question

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.

Answer 1

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 ki logram")`
Moles of ethylene glycol `=(45"g")/(62"g mol"^(-1))=0.73" mol"`
Mass of water in kg `=(600g)/(1000"g kg"^(-1))=0.6" kg"`
Hence molality of ethylene glycol `=(0.73" mol")/(0.60" kg")=1.2" mol kg"^(-1)`
Therefore freezing point depression,
`DeltaT_(f)=1.86" K kg mol"^(-1)xx1.2" mol kg"^(-1)=2.2" K"`
Freezing point of the aqueous solution `=273.15" K"-2.2" K"=270.95" K"`