Solutions Test 12

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Solutions Test 12

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Question 1
1 / -0

Only One Option Correct Type
This section contains 7 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct
Q.
The degree of dissociation (α ) o f weak electrolyte A_xB_y is related to van’t Hoff factor (i) by the expression
[AIEEE 2011]

A

B

C

D

Solution

(a) van’t Hoff factor (i) = [1 + (n - 1)α] where, n = number of ions from one mole of solute
α = degree of ionisation/association
Question 2
1 / -0

0.004 M Na₂S0₄ aqueous solution is isotonic with 0.01 M glucose solution at 300 K. Thus, degree of dissociation of Na₂S0₄ is

A
75 %

B
50 %

C
25 %

D
85 %

Solution

i (van’t Hoff factor) = 1 + (y — 1) x = (1 + 2 x)
∴ π, (osmotic pressure) = MRTi = 0.004 x R x 300 x (1 + 2x)
Glucose is a non-electrolyte.
Hence,
i = 1
π₂ = mRTi = 0.01 x R x 300 x 1 Two solutions are isotonic, hence
Question 3
1 / -0

When 20 g of naphthoic acid (C₁₁H₈O₂) is dissolved in 50 g of benzene, a freezing point depression of 2 K is observed. [K_f (benzene) = 1.72Kmol^-1 kg]. The van’t Hoff factor (i) is
[IIT - JEE 2007]

A
0.5

B
1

C
2

D
3

Solution

w₁| = 20.0 g
m₁ = 11x 12 + 8 x 1 + 2 x 16 = 172 g mol^-1 w₂ = 50.0 g
Question 4
1 / -0

An aqueous solution of a solute AB has boiling point of 101.08° C and freezes at -1 .80 °C . AB is found to be 100% ionised at boiling point. If K_b /K_f = 0.3, then AB

A
is 100% ionised at the freezing point of solution

B
behaves as non-electrolyte at the freezing point of solution

C
forms dimer at the freezing point of solution

D
forms trimer at the freezing point of solution

Solution

Thus, AS behaves as a non-electrolyte at the freezing point of solution (remains non-ionised).
Question 5
1 / -0

Osmolarity of 0.02 M potassium ferrocyanide solution at 300 K is (assume solute is 100% ionised)

A
0.10 osmol

B
0.20 osmol

C
0.02 osmol

D
0.004 osmol

Solution

Osmotic pressure is a colligative property, hence osmolarity is the molarity of the solution involving van't Hoff factor
Thus, osmolarity = 0.02 x i= 0.02 x 5 = 0.I0 osmol
Question 6
1 / -0

AB and CD are two non -volatile solutes dissolved in same solvent, showing following behaviour
AB is ionised in solvent, α = degree of ionisation
CD is dimerised in solvent, 3 = degree of dimerisation
If α = 0.80, then β is

A
0.50

B
0.40

C
0.80

D
0.20

Solution

(d) For A S ,i = 1 + ( y - 1 ) a y = number of particles from one mole of solute = 2
Question 7
1 / -0

Which has the highest boiling point?

A
0.1 m Na₂S0₄

B
0.1 m C₆H₁₂0₆ (glucose)

C
0.1 m MgCI₂

D
0.1 m AI(N0₃)₃

Solution

T_b (boiling point) = T₀(solvent) + ΔT_b
ΔT_b = m x K_b x i
m(same) = 0.1 molal
Δ T_b α i
Greater the value of i, larger the boiling point of solution.