1.

How much amount of KCl must be added to 1 kg of water so that the freezing point is depressed by 2 K? (K_f"for water" = 1.86 kg mol^(-1))

Answer»

40 g
20 g
10 g
60 g

Solution :Since KCl undergoes complete dissociation.
`KCl rarr K^(+) Cl^(-)`
One mole of KCl will give 2 mole particles, therefore, the VALUE of i wil be equal to 2.
`Delta T_f =i K_f m, K_f=1.86 K` kg `mol^(-1)`, `DeltaT_(f) =2 K, i=2`
`therefore 2=2 xx 1.86 xx m`
or `m = (2)/(2 xx 1.86) =0.5376` mol /kg
Grams of KCl `=0.5376 xx 74.5 = 40.05g ` per kg
`therefore` 40.05 g of KCl should be added to 1 kg of WATER.


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