Chemistry Test 177

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Chemistry Test 177

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Weekly Quiz Competition

Question 1
4 / -1

Which one among the following is the correct option for right relationship between C_P and C_V for one mole of ideal gas?

A
C_V = RC_P

B
C_P + C_V = R

C
C_P − C_V = R

D
C_P = RC_V

Solution

At constant volume, q_V = C_VΔT = ΔU

At constant pressure, q_P = C_PΔT = ΔH

For a mole of an ideal gas,

ΔH = ΔU + Δ(PV)

= ΔU + Δ(RT)

= ΔU + RΔT

On putting the values of ΔH and ΔU, we have

C_PΔT = C_VΔT + RΔT

C_P = C_V + R

C_P − C_V = R
Question 2
4 / -1

Which of the following statements is correct?

A
The pH of one litre solution containing 0.49 g of H₂SO₄ is 2.0.

B
The conjugate base of H₂S is S^2-.

C
BF₃ is a Lewis base.

D
Phenolphthalein is colourless in basic medium.

Solution

In 1 L solution of H₂SO₄,

Therefore, [H⁺] = 10^-2

pH = -log 10^-2 = 2

Hence, option (1) is correct.

All the other options are incorrect.

The conjugate base of H₂S is HS^-.

BF₃ is a Lewis acid.

Phenolphthalein is pink in basic medium.
Question 3
4 / -1

When a reversible reaction is at equilibrium, one of its products' concentration is decreased. The equilibrium state

A
Shifts in the direction of backward reaction

B
Shifts in the direction of forward reaction

C
Is not affected

D
Moves in either direction

Solution

According to Le Chatelier's principle, when a system at equilibrium is subjected to a change, the equilibrium shifts in the direction so as to undo the change. Therefore, if the concentration of product is decreased, the forward reaction takes place so as to increase the concentration.
Question 4
4 / -1

An acidic buffer can be prepared by making a solution of

A
formic acid and sodium formate

B
HCI and NaCI

C
NaOH and NaCI

D
NH₄CI and NH₄OH

Solution

An acidic buffer can be prepared by making a solution of an acid and its salt solution. So, an acidic buffer can be prepared by making a solution of formic acid and sodium formate.
Question 5
4 / -1

Which of the following pairs acts as a buffer solution?

A
NaCl + NaOH

B
CH₃COONa + CH₃COOH

C
CH₃COOH + CH₃COONH₄

D
H₂SO₄ + CuSO₄

Solution

Buffer solutions are those solutions which resist the change in pH on addition of small amount of acid or base.

Buffers solutions are the mixture of either a weak acid and its salt with a strong base or a weak base and its salt with a strong acid.

(a) NaCl + NaOH is not a buffer solution because NaOH is a strong base.

(c) CH₃COOH + CH₃COONH₄ is not a buffer solution because CH₃OONH₄ is a salt with a weak base.

(d) H₂SO₄+ CuSO₄ is not a buffer solution because H₂SO₄ is a strong acid.

(d) CH₃COOH + CH₃COONa is a buffer solution because CH₃COOH is a weak acid and CH₃COONa is its salt with a strong base.
Question 6
4 / -1

Two moles of PCl₅ are heated in a closed vessel of 2 L capacity. When the equilibrium is attained, 40% of it is found to be dissociated. What is K_c in mol/dm³?

A
0.532

B
0.267

C
0.133

D
0.174

Solution
Question 7
4 / -1

Which of the following reactions will proceed in forward direction if the volume of the container is increased?

A
H₂ + Cl₂→ 2HCl

B
CaCO₃ → CaO + CO₂

C
N₂ + 3H₂→ 2NH₃

D
N₂ + 2O₂→ N₂O₄

Solution

If the total volume of a container is increased, this means that the total pressure is decreased. The equilibrium will then shift to the side with more moles of gases. In option 2, since CaO and CaCO₃ are solids, we can assume that their concentration are constant. But CO₂ is a gas. Therefore, in this case, the reaction will proceed in forward direction.
Question 8
4 / -1

Among the following hydroxides, the one which has the lowest value of K_sp at ordinary temperature (about 25°C) is

A
Mg(OH)₂

B
Ca(OH)₂

C
Ba(OH)₂

D
Be(OH)₂

Solution

Be(OH)₂ has the lowest value of K_sp at ordinary temperature because Be²⁺ ion is smaller than the other metal ions in the group, which results in a tighter bond with the OH^– ions, thus much lower solubility.

The solubility of a hydroxides of group 2 elements increases down the group because as you go down the group, the size of metal increases, thereby increasing the bond length and decreasing the bond energy.
Question 9
4 / -1

The solubility of M₂S salt is 3.5 × 10^-6. Find out the solubility product.

A
1.7 × 10^-6

B
1.7 × 10^-16

C
1.7 × 10^-18

D
1.7 × 10^-12

Solution

Let 's' be the solubility of salt M₂S which undergoes dissociation as follows:-

Hence, the solubility product (K_sp) = (s) × (2s)²

∴ K_sp = 4 × s³

∴ K_sp = 4 × (3.5 × 10^-6)³

∴ K_sp = 1.7 × 10^-16
Question 10
4 / -1

What is the pH of a 0.10 M solution of barium hydroxide, Ba(OH)₂?

A
1.3

B
11.7

C
13.3

D
None of these

Solution

Ba(OH)₂(s) → Ba²⁺(aq) + 2OH^-(aq)

One mole of Ba(OH)₂ give 2 moles of OH^- ions.

It is given that the concentration of Ba(OH)₂ solution is 0.10 M.

Hence, the concentration of OH^- ions is 0.20 M

pOH = -log[OH^-]

= -log(0.20)

= 0.699

From pOH calcuate the pH of the solution as follows:

pH = 14 - pOH

= 14 - 0.699

= 13.3

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Chemistry Test 177

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Question 1

CV = RCP

CP + CV = R

CP − CV = R

CP = RCV

Solution

Question 2

The pH of one litre solution containing 0.49 g of H2SO4 is 2.0.

The conjugate base of H2S is S2-.

BF3 is a Lewis base.

Phenolphthalein is colourless in basic medium.

Solution

Question 3

Shifts in the direction of backward reaction

Shifts in the direction of forward reaction

Is not affected

Moves in either direction

Solution

Question 4

formic acid and sodium formate

HCI and NaCI

NaOH and NaCI

NH4CI and NH4OH

Solution

Question 5

NaCl + NaOH

CH3COONa + CH3COOH

CH3COOH + CH3COONH4

H2SO4 + CuSO4

Solution

Question 6

0.532

0.267

0.133

0.174

Solution

Question 7

H2 + Cl2 → 2HCl

CaCO3 → CaO + CO2

N2 + 3H2 → 2NH3

N2 + 2O2 → N2O4

Solution

Question 8

Mg(OH)2

Ca(OH)2

Ba(OH)2

Be(OH)2

Solution

Question 9

1.7 × 10-6

1.7 × 10-16

1.7 × 10-18

1.7 × 10-12

Solution

Question 10

1.3

11.7

13.3

None of these

Solution

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C_V = RC_P

C_P + C_V = R

C_P − C_V = R

C_P = RC_V

The pH of one litre solution containing 0.49 g of H₂SO₄ is 2.0.

The conjugate base of H₂S is S^2-.

BF₃ is a Lewis base.

NH₄CI and NH₄OH

CH₃COONa + CH₃COOH

CH₃COOH + CH₃COONH₄

H₂SO₄ + CuSO₄

H₂ + Cl₂→ 2HCl

CaCO₃ → CaO + CO₂

N₂ + 3H₂→ 2NH₃

N₂ + 2O₂→ N₂O₄

Mg(OH)₂

Ca(OH)₂

Ba(OH)₂

Be(OH)₂

1.7 × 10^-6

1.7 × 10^-16

1.7 × 10^-18

1.7 × 10^-12