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Q. No.Determine the solubility of \(Bi_2S_3\) in water, with \(\mathrm{K}_{\mathrm{sp}}\left(\mathrm{Bi}_{2} \mathrm{~S}_{3}\right)=1.08 \times 10^{-73} \text {} \), and express the answer in the form \(x\times 10^{-15} \). What is the value of \(x\)?
1. 3
2. 5
3. 6
4. 1
Subtopic: Solubility Product |
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Level 1: 80%+
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If 200 mL of 0.01 M HCl is mixed with 400 mL of 0.01 M H2SO4, what is the pH of the resulting mixture?
(Given: log 2 = 0.30, log 3 = 0.48, log 5 = 0.70, log 7 = 0.84, log 11 = 1.04)
1. 1.14
2. 1.78
3. 2.34
4. 3.02
(Given: log 2 = 0.30, log 3 = 0.48, log 5 = 0.70, log 7 = 0.84, log 11 = 1.04)
1. 1.14
2. 1.78
3. 2.34
4. 3.02
Subtopic: pH calculation |
74%
Level 2: 60%+
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Given below are two statements:
In light of the above statements choose the correct answer from the options given below:
| Statement I: | According to Lewis's concept, H2O acts as an amphoteric substance. |
| Statement II: | H2O act as an acid on reaction with NH3, while it acts as a base on reaction with H2S. |
| 1. | Both Statement I and Statement II are correct. |
| 2. | Both Statement I and Statement II are incorrect. |
| 3. | Statement I is correct and Statement II is incorrect. |
| 4. | Statement I is incorrect and Statement II is correct. |
Subtopic: Acids & Bases - Definitions & Classification |
Level 4: Below 35%
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When 50 mL, 0.1 M CH3COOH (aq.) is titrated with 25 mL of 0.1 M NaOH (aq) solution. What is the value of the pH of the resulting solution?
[Given: pKa=4.76]
1. 3.76
2. 6.76
3. 4.76
4. 5.76
[Given: pKa=4.76]
1. 3.76
2. 6.76
3. 4.76
4. 5.76
Subtopic: pH calculation | Buffer |
80%
Level 1: 80%+
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The dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})\) is observed at 298 K and 1 atm, with 50% dissociation at equilibrium. What would be the value of the standard Gibbs free energy change, (\(\Delta G^o\))?
| 1. | -684.7 J | 2. | 684.7 J |
| 3. | -342.35 J | 4. | 342.35 J |
Subtopic: Kp, Kc & Factors Affecting them |
60%
Level 2: 60%+
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pH of a buffer solution of \(CH_3CH_2COOH(aq.)~\&~CH_3CH_2COONa(aq.) \) is 4.
Then, the ratio of \(\frac{[CH_3CH_2COO^-]}{[CH_3CH_2COOH]}\) is-
(Given \(K_a=10^{-5} \))
1. 0.1
2. 10
3. 0.2
4. 20
Then, the ratio of \(\frac{[CH_3CH_2COO^-]}{[CH_3CH_2COOH]}\) is-
(Given \(K_a=10^{-5} \))
1. 0.1
2. 10
3. 0.2
4. 20
Subtopic: pH calculation |
70%
Level 2: 60%+
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For an equilibrium reaction :
\(A \rightleftharpoons B+\frac{1}{2} C \)
The correct relation between the degree of dissociation\((\alpha)\), equilibrium pressure(p) and equilibrium constant(\(K_p\)) is :
1. \(K_p =\frac{\alpha^{1 / 2} 2 p^{1 / 2}}{(2+\alpha)^{1 / 2}} \)
2. \(K_p =\frac{\alpha^{1 / 2} p^{3 / 2}}{(2+\alpha)^{3 / 2}}\)
3. \(K_p =\frac{\alpha^{1 / 2} 2 p^{1 / 2}}{(2+\alpha)^{3 / 2}}\)
4. \(K_p =\frac{\alpha^{3 / 2} p^{1 / 2}}{(2+\alpha)^{1 / 2}(1-\alpha)}\)
\(A \rightleftharpoons B+\frac{1}{2} C \)
The correct relation between the degree of dissociation\((\alpha)\), equilibrium pressure(p) and equilibrium constant(\(K_p\)) is :
1. \(K_p =\frac{\alpha^{1 / 2} 2 p^{1 / 2}}{(2+\alpha)^{1 / 2}} \)
2. \(K_p =\frac{\alpha^{1 / 2} p^{3 / 2}}{(2+\alpha)^{3 / 2}}\)
3. \(K_p =\frac{\alpha^{1 / 2} 2 p^{1 / 2}}{(2+\alpha)^{3 / 2}}\)
4. \(K_p =\frac{\alpha^{3 / 2} p^{1 / 2}}{(2+\alpha)^{1 / 2}(1-\alpha)}\)
Subtopic: Introduction To Equilibrium | Kp, Kc & Factors Affecting them |
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Level 2: 60%+
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Subtopic: pH calculation |
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Level 1: 80%+
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Initially 2 moles of \(NOCl \) taken in 1 L of closed container and it dissociates into NO and Cl2 gas:
The reaction is as follows:
\(2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_2(g)\)
Calculate the equilibrium constant \(K_c\) given that 0.4 moles of NO is obtained at equilibrium:
The reaction is as follows:
\(2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_2(g)\)
Calculate the equilibrium constant \(K_c\) given that 0.4 moles of NO is obtained at equilibrium:
| 1. | 142 × 10-4 | 2. | 142 × 10-6 |
| 3. | 125 × 10-4 | 4. | 130 × 104 |
Subtopic: Introduction To Equilibrium |
82%
Level 1: 80%+
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For a hypothetical reaction
A ⇋ B ; Keq = 102
(Use T = 27 °C, R = 8.3 JK−1mol−1, ln10 = 2.3)
If the value of ∆G° for the above reaction is −x kJ, the value of 2x will be:
(Round off to the nearest integer)
1. 11
2. 19
3. 23
4. 28
A ⇋ B ; Keq = 102
(Use T = 27 °C, R = 8.3 JK−1mol−1, ln10 = 2.3)
If the value of ∆G° for the above reaction is −x kJ, the value of 2x will be:
(Round off to the nearest integer)
1. 11
2. 19
3. 23
4. 28
Subtopic: Kp, Kc & Factors Affecting them |
77%
Level 2: 60%+
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