Chemistry Test 99

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Chemistry Test 99

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Weekly Quiz Competition

Question 1
4 / -1

The volume of reaction vessel having following equilibrium increases

SO₂Cl₂(g) ⇌ SO₂(g) + Cl₂(g)

When equilibrium re–established :

A
Amount of SO₂Cl₂ remain unchanged

B
Amount of SO₂ increases

C
Amount of SO₂Cl₂ increases

D
Amount of Cl₂ decreases
Question 2
4 / -1

The exothermic formation of CℓF₃ is represented by the equation :

Cℓ_2(g) + 3F_2(g) ⇌ 2CℓF_3(g) ; ΔH = –329 kJ

Which one of the following will increase the quantity of CℓF₃ in an equilibrium mixture of Cℓ₂, F₂ and CℓF₃.

A
Increasing the temperature

B
Removing Cℓ₂

C
Increasing the volume of container

D
Decreasing the volume of container
Question 3
4 / -1

For the reaction : PCl_5(g) PCl_3(g) + Cl_2(g), the forward reaction at constant temperature is favoured by :-

(a) Introducing inert gas at constant pressure

(b) Introducing inert gas at constant volume

(c) Introducing PCl₅ at constant volume

(d) Introducing Cl₂ at constant volume

A
(a), (b)

B
(a), (c)

C
(b), (c)

D
(a), (d)
Question 4
4 / -1

If 100 mL, 0.1 N H₂SO₄ solution is mixed with 100 mL, 0.1 M Ba(OH)₂ solution then pH of resulting solution will be :

A
1.3

B
12.7

C
7

D
1

Solution

NV = (NV)_Base – (NV)_acid
N(100+100) = 0.2 × 100 – 0.1 × 100
N = 10/200 {Here N = OH^–}
POH = – log [OH^–]

POH = – log 1/20 = 1.3

pH = 14 – 1.3 = 12.7
Question 5
4 / -1

Which of the following solution has pH = 10 at 25°C:-

A
10^–10 M HCl solution

B
10^–10 M NaOH solution

C
10^–4 M NaOH solution

D
Both (1) and (3)
Question 6
4 / -1

Which of the following is correct for poly basic acid :

A
Ka₁ > Ka₂

B
Ka₂ > Ka₁

C
Ka₁ = Ka₂

D
Ka₁ > 1/Ka₂
Question 7
4 / -1

Nicotinic acid (K_a = 1.4 × 10^–5) is represented by formula HNiC. It's percentage dissociation in a solution containing 0.1 Mole nicotinic acid in 2L solution will be :-

A
2.82%

B
3.34%

C
1.67%

D
2.14%
Question 8
4 / -1

The [OH^–] in 0.01 M aqueous solution of NaOCN (K_b for OCN^– = 10^–10) will be :-

A
10^–6 M

B
10^–3 M

C
10^–8 M

D
10^–10 M

Solution

For NaOCN (WASB) solution
Question 9
4 / -1

The solubility of BaSO₄ at 25°C is 2.33 ×10^–4 gm/100 ml. Find out the solubility product of BaSO₄
(M.wt. of BaSO₄ = 233) :

A
1 × 10^–8

B
1 × 10^–5

C
2.33 × 10^–8

D
1 × 10^–10
Question 10
4 / -1

What is the molar solubility of Ag₂CO₃ (K_sp = 4 × 10^–13) in 0.1 M Na₂CO₃ solution ?

A
10^–6 M

B
10^–7 M

C
2 × 10^–6 M

D
2 × 10^–7 M

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Chemistry Test 99

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Chemistry Test 99

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Question 1

Amount of SO2Cl2 remain unchanged

Amount of SO2 increases

Amount of SO2Cl2 increases

Amount of Cl2 decreases

Question 2

Increasing the temperature

Removing Cℓ2

Increasing the volume of container

Decreasing the volume of container

Question 3

(a), (b)

(a), (c)

(b), (c)

(a), (d)

Question 4

1.3

12.7

7

1

Solution

Question 5

10–10 M HCl solution

10–10 M NaOH solution

10–4 M NaOH solution

Both (1) and (3)

Question 6

Ka1 > Ka2

Ka2 > Ka1

Ka1 = Ka2

Ka1 > 1/Ka2

Question 7

2.82%

3.34%

1.67%

2.14%

Question 8

10–6 M

10–3 M

10–8 M

10–10 M

Solution

Question 9

1 × 10–8

1 × 10–5

2.33 × 10–8

1 × 10–10

Question 10

10–6 M

10–7 M

2 × 10–6 M

2 × 10–7 M

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Amount of SO₂Cl₂ remain unchanged

Amount of SO₂ increases

Amount of SO₂Cl₂ increases

Amount of Cl₂ decreases

Removing Cℓ₂

10^–10 M HCl solution

10^–10 M NaOH solution

10^–4 M NaOH solution

Ka₁ > Ka₂

Ka₂ > Ka₁

Ka₁ = Ka₂

Ka₁ > 1/Ka₂

10^–6 M

10^–3 M

10^–8 M

10^–10 M

1 × 10^–8

1 × 10^–5

2.33 × 10^–8

1 × 10^–10

10^–6 M

10^–7 M

2 × 10^–6 M

2 × 10^–7 M