Question

Which of the following solutions would have the highest boiling point?   

0.10 m NaCl

0.10 m CaCl2

0.10 m Na2SO4

0.10 m of sucrose (C12H11O22)

0.10 m Na3PO4

Answer 1

ΔTb = i*Kb*m

ΔTb is the elevation in boiling point

ΔTb will be maximum when i*m is maximum

so, higher boiling point is for that solution which has higher value of i*m

NaCl dissociates into 1 Na+ and 1 Cl-

So, i = 2

Hence, i*m = 2*0.1 = 0.2

CaCl2 dissociates into 1 Ca2+ and 2 Cl-

So, i = 3

Hence, i*m = 3*0.1 = 0.30

Na2SO4 dissociates into 2 Na+ and 1 SO42-

So, i = 3

Hence, i*m = 3*0.1 = 0.30

C12H22O11 is non electrolyte and hence it will not dissociate

So, i = 1

Hence, i*m = 1*0.1 = 0.1

Na3PO4 dissociates into 3 Na+ and 1 PO43-

So, i = 4

Hence, i*m = 4*0.1 = 0.4

Decreasing order of i*m is:

Na3PO4

Na2SO4

CaCl2

NaCl

C12H22O11

This will be the order of decreasing boiling point

So, compounds from higher to lower boiling point are:

Na3PO4

Na2SO4

CaCl2

NaCl

C12H22O11

Answer: Na3PO4