Chemistry Test 178

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Chemistry Test 178

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Weekly Quiz Competition

Question 1
4 / -1

The solubility of CaF₂ in pure water is 2.3 × 10^-4 mol dm^-3. Its solubility product will be

A
4.6 × 10^-4

B
4.6 × 10^-8

C
6.9 × 10^-12

D
4.9 × 10^-11

Solution

Solubility product = 4S³

= 4(2.3 × 10^-4)³

= 4 × 12.167 × 10^-12

= 48.67 × 10^-12

= 4.867 × 10^-11 = 4.9 × 10^-11
Question 2
4 / -1

The solubility of Mg(OH)₂is S moles/litre. The solubility product under the same conditions is

A
4S³

B
3S⁴

C
4S²

D
S³

Solution

K_sp = [Mg²⁺] [OH^-]²

K_sp = S × (2S)²

K_sp = 4S³
Question 3
4 / -1

In the reaction 2H₂O (g) + 2Cl₂(g) ⇄ 4HCl (g) + O₂(g), the values of K_p and K_c are related as

A
K_p = K_c

B
K_p> K_c

C
K_p < K_c

D
K_p = K_c = 0

Solution

Therefore, K_P > K_C for those reactions that proceed with an increase in the number of gaseous moles.

2H₂O (g) + 2Cl₂(g) ⇄ 4HCl (g) + O₂(g)

K_p = K_cRT

So, K_p> K_c
Question 4
4 / -1

If the concentration of [Ag⁺] = 10^-5 in a solution, then the concentration of [Cl^-] required for the precipitation of AgCl (Ksp = 2 ×10^-12) is

A
2 × 10^-5

B
2 × 10^-7

C
1 × 10^-8

D
1 × 10^-9

Solution

Use the expression: K_sp= [Ag⁺][Cl^-]

2 × 10^-12 = 10^-5 × [Cl^-]

[Cl^-] = 2 × 10^-7
Question 5
4 / -1

Which of the following compounds is not a protonic acid?

A
SO(OH)₂

B
SO₂(OH)₂

C
B(OH)₃

D
PO(OH)₃

Solution

Since B(OH)₃is not able to donate any proton, so it is not a protonic acid.
Question 6
4 / -1

For a sparingly soluble salt A_pB_q, the relationship between its solubility product (L_s) and its solubility (S) is

A
L_s = S^p+qp^p q^q

B
L_s = S^p+q p^q q^p

C
L_s= S^pq p^p q^p

D
L_s = S^pq (pq)^q+p

Solution

A_pB_q = pA⁺ + qB^-

If S is the solubility,

[A⁺] = pS, [B^-] = qS;

K_sp = [A⁺]^p[B^-]^q

L_s= (pS)^p (qS)^q = S^p+qp^p q^q
Question 7
4 / -1

What is the maximum concentration of equimolar solution of ferrous sulphate and sodium sulphide, so that when mixed in equal volumes there is no precipitation of iron sulphide? [K_sp for iron sulphide = 6.3 × 10^-18]

A
4 × 10^-9mol L^-1

B
4.8 × 10^-11mol L^-1

C
5.02 × 10^-9mol L^-1

D
6 × 10^-9mol L^-1

Solution

Let the concentration of FeSO₄ and Na₂S be x mol L^-1.

Then, after mixing equal volumes:

[FeSO₄] = [Na₂S] = x/2 M

Or [Fe²⁺] = [S^2-] = x/2 M

FeS ⇌ Fe²⁺ + S^2-

K_sp = [Fe²⁺][S^2-]

6.3 × 10^-18 = (x/2)(x/2)

On solving, x = 5.02 × 10^-9 mol L^-1
Question 8
4 / -1

The solubility product of silver bromide is 5 x 10^-13. The quantity of potassium bromide with molar mass 120 g/mol to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is

A
5 x 10^-8 g

B
1.2 x 10^-10 g

C
1.2 x 10^-9 g

D
6.2 x 10^-5 g

Solution
Question 9
4 / -1

For the equilibrium A ⇋ B, the variation of the rate of the forward (a) and reverse (b) reaction with time is given by

A

B

C

D

Solution
Question 10
4 / -1

The number of H⁺ ions in 1 cm³ of a solution having pH = 13 is ___________.

A
6.023 × 10⁴

B
6.023 × 10⁷

C
6.023 × 10¹⁰

D
6.023 × 10¹³

Solution

For pH = 13,

[H⁺] = 10^-13 M

Number of moles H⁺ ions in 1000 cm³(1 L)solution = 10^-13

Number of moles of H⁺ ions in 1 cm³ solution

Number of moles of H⁺ ions in 1 cm³ solution = Number of moles × 6.023 × 10²³

Number of moles of H⁺ ions in 1 cm³ solution = 10^-16 × 6.023 10²³ = 6.023 × 10⁷

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Chemistry Test 178

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Chemistry Test 178

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Question 1

4.6 × 10-4

4.6 × 10-8

6.9 × 10-12

4.9 × 10-11

Solution

Question 2

4S3

3S4

4S2

S3

Solution

Question 3

Kp = Kc

Kp > Kc

Kp < Kc

Kp = Kc = 0

Solution

Question 4

2 × 10-5

2 × 10-7

1 × 10-8

1 × 10-9

Solution

Question 5

SO(OH)2

SO2(OH)2

B(OH)3

PO(OH)3

Solution

Question 6

Ls = Sp+q pp qq

Ls = Sp+q pq qp

Ls = Spq pp qp

Ls = Spq (pq)q+p

Solution

Question 7

4 × 10-9 mol L-1

4.8 × 10-11 mol L-1

5.02 × 10-9 mol L-1

6 × 10-9 mol L-1

Solution

Question 8

5 x 10-8 g

1.2 x 10-10 g

1.2 x 10-9 g

6.2 x 10-5 g

Solution

Question 9

Solution

Question 10

6.023 × 104

6.023 × 107

6.023 × 1010

6.023 × 1013

Solution

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4.6 × 10^-4

4.6 × 10^-8

6.9 × 10^-12

4.9 × 10^-11

4S³

3S⁴

4S²

S³

K_p = K_c

K_p> K_c

K_p < K_c

K_p = K_c = 0

2 × 10^-5

2 × 10^-7

1 × 10^-8

1 × 10^-9

SO(OH)₂

SO₂(OH)₂

B(OH)₃

PO(OH)₃

L_s = S^p+qp^p q^q

L_s = S^p+q p^q q^p

L_s= S^pq p^p q^p

L_s = S^pq (pq)^q+p

4 × 10^-9mol L^-1

4.8 × 10^-11mol L^-1

5.02 × 10^-9mol L^-1

6 × 10^-9mol L^-1

5 x 10^-8 g

1.2 x 10^-10 g

1.2 x 10^-9 g

6.2 x 10^-5 g

6.023 × 10⁴

6.023 × 10⁷

6.023 × 10¹⁰

6.023 × 10¹³