Calculate the amount of KCl which must be added to 1 kg of water so that its freezing point is depressed by 2 K.

Explanation

`KClrarrK^(+)+Cl^(-)`, (Here i = 2)
`DeltaT_(f)=iK_(f)m,2=2xx1.86xxm`
`rArrm=2/(2xx1.86)=0.54`
Hence, amount pf KCl to be dissolved in 1 kg of water = 0.54 mol = `0.54xx74.5=40.23g`