Question
Which has the greater boiling point elevation: 3 g Ca(NO3)2 in 60 g water or 6 g Ca(NO3)2 in 30 g water?

Which will have a small freezing point depression: 100 g of methanol (CH3OH) in 500 g of water or 200 g of methanol in 250 g of water?

Answer:

Moles Ca(NO3)2 = 3 / 164 g/mol =0.018
molality = 0.018 mol / 0.060 Kg = 0.3
Ca(NO3)2 >> Ca2+ + NO32- => i = 3
delta T = i x kb x m = 3 x 0.512 x 0.3 = 0.46
b.p. = 100.46 °C

Moles Ca(NO3)2 = 6 / 164 =0.037
molality = 0.037 / 0.030 Kg =1.2
delta T = 3 x 0.512 x 1.2 = 1.8
b.p. = 101.8 °C

Moles CH3OH = 100 g/ 32 g/mol =3.1
m = 3.1 / 0.500 Kg =6.2
delta T = kf x m = 1.86 x 6.2 =11.5
f.p. = – 11.5

Moles CH3OH = 200 / 32 = 6.2
m = 6.2 / 0.250 =24.8
delta T = 1.86 x 24.8 = 46
f.p. = – 46 °C