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1

AIPMT 2004

MCQ (Single Correct Answer)
The solubility product of a sparingly soluble salt AX2 is 3.2 $$ \times $$ 10$$-$$11. Its solubility (in moles/L) is
A
5.6 $$ \times $$ 10$$-$$6
B
3.1 $$ \times $$ 10$$-$$4
C
2 $$ \times $$ 10$$-$$4
D
4 $$ \times $$ 10$$-$$4

Explanation

AX2 A2+ + 2X-
s s 2s


Ksp = = [A2+] [X ]2 = s × (2s)2 = 4s3

$$ \Rightarrow $$ 3.2 $$ \times $$ 10$$-$$11 = 4s3

$$ \Rightarrow $$ s3 = 8 × 10–12

$$ \Rightarrow $$ s = 2 × 10–4 mol L–1
2

AIPMT 2003

MCQ (Single Correct Answer)
The solubility product of AgI at 25oC is 1.0 $$ \times $$ 10$$-$$16 mol2 L$$-$$2. The solubility of AgI in 10$$-$$4 N solution of KI at 25oC is approximately (in mol L$$-$$1
A
1.0 $$ \times $$ 10$$-$$16
B
1.0 $$ \times $$ 10$$-$$12
C
1.0 $$ \times $$ 10$$-$$10
D
1.0 $$ \times $$ 10$$-$$8

Explanation

AgI Ag+ + I-
s s s


Ksp = s2

$$ \Rightarrow $$ 1.0 × 10–16 = s2

$$ \Rightarrow $$ s = 1.0 × 10–8 mol L–1

$$ \therefore $$ [Ag+] = 1.0 × 10–8 mol L–1

Also, in 10–4 N KI solution,

[I–1] = (10–4 + 1.0 × 10–8) mol L–1

$$ \Rightarrow $$ [I–1] = (10–4) mol L–1

[As 1.0 × 10–8 mol L–1 << 1.0 × 10–4 mol L–1]

$$ \therefore $$ Ksp of AgI = [Ag+][I]

= (1.0 × 10–8)(10–4)

= 1.0 × 10–12 mol L–1
3

AIPMT 2003

MCQ (Single Correct Answer)
The reaction quotient (Q) for the reaction

N2(g) + 3H2(g) $$\rightleftharpoons$$ 2NH3(g) is given by

$$Q = {{{{\left[ {N{H_3}} \right]}^2}} \over {\left[ {{N_2}} \right]{{\left[ {{H_2}} \right]}^3}}}$$.

The reaction will proceed from right to left if
A
Q = Kc
B
Q < Kc
C
Q > Kc
D
Q = 0

Explanation

For any reaction to process in forward direction the reaction quotient (Q) must be less than equilibrium constant KC.

Q < Kc
4

AIPMT 2003

MCQ (Single Correct Answer)
In Haber process 30 litres of dihydrogen and 30 litres of dinitrogen were taken for reaction which yielded only 50% of the expected product. What will be the composition of gaseous mixture underthe aforesaid condition in the end ?
A
20 litres ammonia, 20 litres nitrogen, 20 litres hydrogen
B
10 litres ammonia, 25 litres nitrogen, 15 litres hydrogen
C
20 litres ammonia, 10 litres nitrogen, 30 litres hydrogen
D
20 litres ammonia, 25 litres nitrogen, 15 litres hydrogen

Explanation

Balanced chemical equation for Haber's process is as follows :

3H2 + N2 $$ \to $$ 2NH3

It is given that only 50% of the expected product is formed hence only 10 litre of NH3 is formed.

Therefore, composition of gaseous mixture at the end is as follows :

N2 used = 5 litres

N2 left = 30 L – 5 L = 25 L

H2 used = 15 litres,

H2 legt = 30 L – 15 L = 15 L

NH3 = 10 L

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