Question

Match the following aqueous solutions with the appropriate letter from the column on the right.

	1.	0.11 m	Fe(NO3)3		A.	Highest boiling point
	2.	0.21 m	NaI		B.	Second highest boiling point
	3.	0.13 m	Cr(NO3)3		C.	Third highest boiling point
	4.	0.48 m	Glucose(nonelectrolyte)		D.	Lowest boiling point

Answer 1

Ans:

Elevation in Boiling point is determined by the equation:

deltaT = i . Kb . m

[where; deltaT - elevation in boiling point

i - vant hoffs factor(number of particles in solution)

m - molality of the solution

Kb - elevation in boiling point constant]

For water: Kb = 0.512 ºC · kg H2O / mol particles

1. 0.11m Fe(NO₃)₃ :-

Fe(NO₃)₃ dissociates into total 4 ions in solution as shown below:

Fe(NO₃)₃Fe³⁺ + 3NO₃^-

hence, i = 4 ; m = 0.11

deltaT = i . Kb .m = 4 x 0.11 = 0.44Kb

2. 0.21m NaI : -

NaI Na⁺ + I^-

Here i = 2 and m = 0.21

hence, delta T = i . Kb. m = 2 x 0.21 = 0.42Kb

3. 0.13m Cr(NO3)3

Cr(NO₃)₃ Cr³⁺ + 3NO₃^-

here, i = 4, m = 0.13

deltaT = 4 x Kb x 0.13 = 0.52Kb

4. 0.48m glucose:

since it is a non electrolyte it doesn't dissociate in solution,

hence i = 1 (1 glucose molecule)

deltaT = i . Kb . m = 1 x 0.48 = 0.48Kb

The one with highest elevation in boiling point has the highest boiling point.

deltaT = Tsolution - 100 deg C

Tsolution = 100 + deltaT

It can be summarised as below:

S.No:	Substance	i	deltaT = i.kb.m	deltaT Elevation in boiling point	deltaT	Tsolution
1.	0.11m Fe(NO3)3	4	4 x 0.11	0.44Kb	0.22528	100.225	C.	Third highest boiling point
2.	0.21m NaI	2	2 x 0.21	0.42Kb	0.21504	100.215	D.	Lowest boiling point
3.	0.13m Cr(NO3)3	4	4 x 0.13	0.52Kb	0.26624	100.266	A.	Highest boiling point
4.	0.48m Glucose(nonelectrolyte)	1	1 x 0.48	0.48Kb	0.24576	100.246	B.	Second highest boiling point