Calculate the amount of `KCl` which must be added to `1kg` of water so that the freezing point is depressed by `2K`. (`K_(f)` for water `=1.86K kg "mol"^(-1)`).

Answer

`40.05 %`

Explanation

Since KCl undergoes complete dissociaiton as :
`KCl to K^(+) + CL^(-)`
One mole of KCl will give 2 mole particle and the value of .I.will be equal to 2.
`Delta T = ik_(f) m`
`k_(f) = 1.86 K kh mol^(-1) , Delta T_(f) = 2k, i=2`
`therefore " "2 = 2 xx 1.86 xx m`
or `" " m = (2)/(2 xx 1.86) = 0.5376 mol//kg`
Grams of KCl of `=0.537 xx 74.5 = 40.05` g per kg.