Past JEE Questions Ch.9 SOLUTIONS

JEE Question 2007 paper I

When 20 g of naphthoic acid (C-11H-8O-2) is dissolved in 50 g of benzene (K-f = 1.72 K kg mol^-1), a freezing point depression of 2 K is observed. The van’t Hoff factor (i) is


(A) 0.5
(B) 1
(C) 2
(D) 3

Answer: A

Van't Hoff factor = Normal Molar Mass/Observed Molar mass

Observed Molar Mass M-B = K-f*1000*W-B/(W-A*ΔT-f)

K-f = Molal depression constant
W-A = weight of solvent
W-B = weight of solute
ΔT-f = depression in freezing point.

In the problem Normal Molar mass = 172
Observe molar mass = 1.72*1000*20/50*2 = 17.2*20 = 172*2

Van't Hoff factor =172/172*2 = 0.5

Ref: Dr. Jauhar's book unit 3: solutions.