Question

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

	1.	0.18 m	MgI2		A.	Highest boiling point
	2.	0.19 m	Fe(NO3)2		B.	Second highest boiling point
	3.	0.16 m	CrCl2		C.	Third highest boiling point
	4.	0.50 m	Sucrose(nonelectrolyte)		D.	Lowest boiling point

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

	1.	0.13 m	AlBr3		A.	Lowest freezing point
	2.	0.24 m	CoSO4		B.	Second lowest freezing point
	3.	0.17 m	Ba(NO3)2		C.	Third lowest freezing point
	4.	0.56 m	Glucose(nonelectrolyte)		D.	Highest freezing point

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

	1.	0.18 m	MgI2		A.	Lowest freezing point
	2.	0.19 m	Fe(NO3)2		B.	Second lowest freezing point
	3.	0.16 m	CrCl2		C.	Third lowest freezing point
	4.	0.50 m	Sucrose(nonelectrolyte)		D.	Highest freezing point

Answer 1

Answer

1)

1. 0.18m MgI2 : B second highest boiling point

2. 0.19m Fe(NO3)2 : A highest boiling point

3. 0.16m CrCl2 : D Lowest boiling point

4. 0.50m sucrose(nonelectrolyte) : C Third highest boiling point

2)

1. 0.13m AlBr3 : B second lowest freezing point

2. 0.24m CoSO4 : A highest freezing point

3.0.17m Ba(NO3)2 : C third lowest freezing point

4. 0.56m Glucose : D lowest freezing point

3)

1. 0.18m MgI2 : B Second lowest freezing point

2. 0.19m Fe(NO3)2 : A Lowest freezing point

3. 0.50m Sucrose (nonelectrolyte) : C Third lowest freezing point

4. 0.16m CrCl2 : D Highest freezing point

Explanation

i) Freezing point depression is calculated by the followning formula

∆T_f = K_f × b × i

where ,

K_f = freezing point depression constant which depends on solvent

b = molality of solute

i = van't Hoff factor

for same solvent , ∆T_f is depends on b×i

if b×i is higher freezing point depression would be higher and freezing point would be lower

For example consider question 2

i for AlBr3 = 4 ( because 4 ions after dissociation)

i for CoSO4 = 2 ( because 2 ions after dissociation )

i for Ba(NO3)2 = 3 ( because 3 ions after dissociation )

i for Glucose = 1 ( because no dissociation)

b × i for AlBr3 = 4× 0.13m = 0.52m

b× i for CoSO4 = 2 × 0.24m = 0.48m

b × i for Ba(NO3)2 = 3 × 0.17 = 0.51 m

b × i for Glucose = 1 × 0.56m = 0.56m

For glucose b× i value is high , so it would have highest freezing point depression and lowest freezing point

ii) Boiling point elevation is calculated by the following dormula

∆T_b = K_b × b × i

where,

K_b = boiling point elevation constant which depends on solvent

Therefore, for same solvent ∆T_b depends on b× i

If b× i is higher boiling point elevation will be higher and boiling point would higher