Calculate the amount of `CaCI_(2) ("molar mass"=111 g mol^(-1))` which must be added to 500 g of water to lower its freezing point by 2 K assuming `CaCI_(2)` to be completely dissoviated `(K_(f) for = 1.86 "K kg mol"^(-1))`.

Explanation

`CaCl_(2)` undergoes complete dissociation as :
`CaCl_(2) to Ca^(2+) + 2Cl^(-)`
One mole of `CaCl_(2)` will give 3 mole particlues and therefore , the vlaue of .I. will be equal to 3.
`Delta T_(f) = iK_(f) xx m`
` = (ixx K _(f) xx w_(B) xx 1000)/(M_(B) xx w_(A))`
`k_(f) = 1.86 K kg mol^(-1) , w_(A) = 500 g, w_(B) = ? , Delta T_(f) = 2K, i=3` ,`M_(B) = 111 g mol^(-1)` .
` 2 = (3 xx 1.86 xx w_(B) xx 1000)/(111 xx 500)`
`therefore " "w_(B) = (2xx 111 xx 500)/(3 xx 1.86 xx 1000) = 19.89 g`