Question

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

	1.	0.25 m	NaNO3		A.	Lowest freezing point
	2.	0.22 m	KOH		B.	Second lowest freezing point
	3.	0.16 m	Pb(NO3)2		C.	Third lowest freezing point
	4.	0.41 m	Glucose(nonelectrolyte)		D.	Highest freezing point

2)

	1.	0.22 m	CuSO4		A.	Lowest freezing point
	2.	0.14 m	MgI2		B.	Second lowest freezing point
	3.	0.16 m	K2CO3		C.	Third lowest freezing point
	4.	0.50 m	Ethylene glycol(nonelectrolyte)		D.	Highest freezing point

3)

	1.	0.12 m	AlI3		A.	Lowest freezing point
	2.	0.11 m	Cr(NO3)3		B.	Second lowest freezing point
	3.	0.14 m	Na2SO4		C.	Third lowest freezing point
	4.	0.50 m	Ethylene glycol(nonelectrolyte)		D.	Highest freezing point

Answer 1

ΔTf = i*Kf*m

ΔTf is the depression in freezing point

ΔTf will be maximum when i*m is maximum

so,

freezing point will be lowest when i*m is maximum

freezing temperature is maximum when ΔTf will be minimum

i is the number of ions that a compound can break into

1)

1)

NaNO3

i = 2

NaNO3 breaks into 1 Na+ ion and 1 NO3- ions

so,

i*m = 2*0.25 = 0.50

2)

KOH

i = 2

so,

i*m = 2*0.22 = 0.44

3)

Pb(NO3)2

similarly:

i = 3

i*m = 3*0.16 = 0.48

4)

glucose

This is non electrolyte

i = 1

i*m = 0.41

1. matches to A

2. matches to C

3. matches to B

4. matches to D

All other questions can be done in same way.

I am allowed to answer only 1 question at a time