What is the freezing point of solution prepared by disolving 15 g of KBr in 100 ml of water? The freezing point depression constant of water is 1.86.
ΔTf=Kf×Cm
Kf(H2O)=1.86
Cm (molality) is number of moles of solute per kilogram solvent.
ΔTf=Kf×m
M (KBr) = 119 g/mol
n (NaCl) = 15/119 = 0.13 mol
Density of water is 1 kg/l. Threfore, mass of 100 ml = 0.1 L of water is 0.1 kg.
Сm (KBr) = 0.13/0.1=1.3 mol
ΔTf=1.86×1.3=2.4
Therefore Tf of an aqueous solution containing 15 g of KBr in 100 ml of H2O will be -2.4°C
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