Question

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

	1.	0.14 m	CaCl2		A.	Lowest freezing point
	2.	9.3×10-2 m	K3PO4		B.	Second lowest freezing point
	3.	8.0×10-2 m	Al2(SO4)3		C.	Third lowest freezing point
	4.	0.34 m	Glucose(nonelectrolyte)		D.	Highest freezing point

Answer 1

Depression in freezing point = i x Kf x m

i= vant hoff's factor = number of ions present in a solute

m= molality

Kf = molal freezing constant

for CaCl2

CaCl2 -------------- Ca+2 + 2 Cl-

So, it contains three ions. i,e i=3

Kf of water = 1.86 C/m

for CaCl2 = 3x1.86x0.14 = 0.7812C

for K3PO4

K3PO4 -------------- 3 K+ + PO4-

i= 4

for K3PO4 = 4x1.86x 9.3x10^-2 = 0.6919 C

for Al2(SO4)3

Al2(SO4)3 ----------------- 2 Al+3 + 3 SO4-2

i= 5

for Al3(SO4)3 = 5 x 1.86 x 8.0x10^-2 = 0.744 C

Glucose

C6H12O6 donot undergoes dissoiciation.So, for this i=`1

for Glucose = 1x1.86x0.34 = 0.6324 C

The freezing point of the solution is lowered by the incresing in the number of moles solute paticles . but different in the molality of the solution.

lowest freezing point is CaCl2 (A)

second lowest is Al2(SO4)3 ( C)

third lowest is K3PO4(B)

Highest freezing point is Glucose (D)