1. a solution contains 8.84 g of H2SO4 in 10 ml and the density is 1.8 g/cc, the molality of the solution is
a) 12.55m
b) 6.20
c) 36.05
d) 9.85
Showing posts with label Solutions. Show all posts
Showing posts with label Solutions. Show all posts
Monday, December 24, 2007
Monday, December 17, 2007
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.
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.
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