Calculate the amount of `CaCl_2` [molar mass = 111 g `"mol"^(-1)` ] which must be added to 500 g of water to lower the freezing point by 2 K, assuming `CaCl_2` is completely dissociated.

Explanation

Apply the relation (without taking dissociation into consideration)
` Delta T_f = (K_f xx 1000 xx w_2)/(M_2 xx w_1)`
`2K = (1.86 K kg "mol"^(-1) xx 1000 g kg^(-1) xx w_2)/(111 g "mol"^(-1) xx 500 g) `
` w_2 = (2 xx 111 xx 500)/(1.86 xx 1000) = 59.68 g`
One molecule of `CaCl_2` dissociates into one `Ca^(2+)` ion and two `Cl^-` ions.
`CaCl_2 to Ca^(2+) + 2Cl^(-)`
particle dissociates into three particles.
Therefore, the amount of `CaCl_2` needed = `(59.68)/(3) = 19.89 g`