Calculate the amount of `CaCl_(2)` (molar mass = 111g `"mol"^(-1)`) which must be added to 500 g of water of lower its freezing point by 2K, assuming `CaCl_(2)` is completely dissociated.
(Kf for water = 1.86 k kg `"mol"^(-1)`).

Explanation

`CaCl_(2)rarrCa^(2+)+2Cl^(Theta)`
One mole of `CaCl_(2)` gives 3 mol of particles and
`thereforei=3,DeltaT_(f)=ik_(f)m`
`DeltaT_(f)=(ikfxxWBxx1000)/(MBxxWA),2=(3xx1.86xxWBxx1000)/(111xx500),W_(B)=(2xx111xx500)/(3xx1.86xx1000)=19.89g`