Question

a) if aqueous solutions were made with the following molal concentrations, which one would have the highest boiling point?

b) which one would have the highest freezing point?

0.15 m Na3PO4

0.15 m Na2SO4

0.25 m MgCl2

0.15 m CaCl2

0.20 m Glucose (non electrolyte)

Answer 1

0.15 m Na3PO4

i=4. So, i*m = 4*0.15 = 0.60

0.15 m Na2SO4

i=3. So, i*m = 3*0.15 = 0.45

0.25 m MgCl2

i=3. So, i*m = 3*0.35 = 0.75

0.15 m CaCl2

i=3. So, i*m = 3*0.15 = 0.45

0.20 m Glucose (non electrolyte)

i=1. So, i*m = 1*0.20 = 0.20

a)

ΔTb = i*Kb*m

ΔTb is the increase in boiling point

ΔTb will be maximum when i*m is maximum

so, highest boiling point if for that solution which has highest value of i*m

Answer: 0.25 m MgCl2

b)

ΔTf = i*Kf*m

ΔTf is the depression in freezing point

ΔTf will be maximum when i*m is maximum

so, highest freezing point if for that solution which has lesser value of i*m

Answer:0.20 m Glucose