1.

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»

Solution :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


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