`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 solution. Given`K_(f)=1.86 K kg mol^(-1)`.

`45 g` of ethylene glycol `C_(2)H_(6)O_(2)` is mixed with `600 g` of w

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 solution. Given`K_(f)=1.86 K kg mol^(-1)`.
Answer» Depression in freezing point is realed to the molality, therefore the molality of the solution with respect to ethylene glycol, `DeltaT_(f)=K_(f)m` Mole of ethylene glycol = `45 gxx(1 mol)/(62g) =0.73 mol` Mass of water in kg =`600 xx (1 kg)/(1000 g)=0.60 kg` Hence, molality of ethylene glycol `=(0.73 mol)/(0.60 kg)=1.20 mol kg` Therefore,freezing point depression `(DeltaT_(f)=1.86 K kg mol^(-1))xx(1.2 mol kg^(-1) )=2.2 K` Freezing point of the aqueous solution `T_(f) =T_(f)^(@) - Delta_(f)T` `273.15 K -2.2 K =270.95 K`

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`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 solution. Given`K_(f)=1.86 K kg mol^(-1)`.

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