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How much amount of KCl must be added to ...

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))`

A

40 g

B

20 g

C

10 g

D

60 g

Text Solution

Verified by Experts

The correct Answer is:
A
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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