Question

Assuming complete dissociation of the solute, how many grams of KNO3 must be added to 275 mL of water to produce a solution that freezes at −14.5 ∘C? The freezing point for pure water is 0.0 ∘C and Kf is equal to 1.86 ∘C/m.

Answer 1

Freezing point depression= kf* molality

14.5 = 1.86* molality

Molality= 14.5/1.86 =7.795

moles of solute KNO3/ kg of water

volume of water= 275 ml

mass of water= 257*1 g/cc) =275 gms=0.275 Kg

from molality information 1 kg of solvent water contaisn 7.795 moles of KNO3

0.275 kg water contaisn 7.795*0.275= 2.143 moles of KNO3

Atomic weights , K= 39 N=14 and O= 16

Hence molecular weight of KNO3= 39+14+3*16= 101

Moles= mass/ molecular weight

mass of KNO3= 2.143*101=210.443 gms