Q. No.For the reaction:
\(\begin{array}{ll} \mathrm{A}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{~A}(\mathrm{~g}) \\ \Delta \mathrm{G}_{\mathrm{f}\left(\mathrm{~A}_2\right)}^{\circ}=-100 \mathrm{KJ} / \mathrm{mol} \\ & \\ \Delta\mathrm{G}_{\mathrm{f}(\mathrm{~A})}^{\circ}=-50.8625 \mathrm{KJ} / \mathrm{mol} \end{array}\)
At 300K and 1 atm, degree of dissociation of \(A_2\) gas at equilibrium is \(x\times 10^{-2}.\) Find x:\(\left[\mathrm{R}=8.3 ~\mathrm{Jmol}^{-1} \mathrm{~K}^{-1}\right]\)
1. 52.70
2. 55.18
3. 57.73
4. 59.30
\(\begin{array}{ll} \mathrm{A}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{~A}(\mathrm{~g}) \\ \Delta \mathrm{G}_{\mathrm{f}\left(\mathrm{~A}_2\right)}^{\circ}=-100 \mathrm{KJ} / \mathrm{mol} \\ & \\ \Delta\mathrm{G}_{\mathrm{f}(\mathrm{~A})}^{\circ}=-50.8625 \mathrm{KJ} / \mathrm{mol} \end{array}\)
At 300K and 1 atm, degree of dissociation of \(A_2\) gas at equilibrium is \(x\times 10^{-2}.\) Find x:\(\left[\mathrm{R}=8.3 ~\mathrm{Jmol}^{-1} \mathrm{~K}^{-1}\right]\)
1. 52.70
2. 55.18
3. 57.73
4. 59.30
Subtopic: Ionisation Constant of Acid, Base & Salt | Solubility Product |
Level 3: 35%-60%
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\(\text K_{sp}\) for Cr(OH)3 is 1.6 × 10–30. What is the molar solubility of this salt in water?
| 1. | \(\sqrt[4]{\dfrac{1.6 \times 10^{-30}}{27}}\) | 2. | \(\dfrac{1.8 \times 10^{-30}}{27}\) |
| 3. | \(\sqrt[5]{1.8 \times 10^{-30}} \) | 4. | \(\sqrt[2]{1.6 \times 10^{-30}}\) |
Subtopic: Solubility Product |
84%
Level 1: 80%+
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Which of the following happens when NH4OH is added gradually to the solution containing 1M A2+ and 1M B3+ ions?
Given: \(\mathrm{K}_{\mathrm{sp}}\left[\mathrm{~A}(\mathrm{OH})_2\right]=9 \times 10^{-10} \text {}\)
\(\mathrm{K}_{\mathrm{sp}}\left[\mathrm{~B}(\mathrm{OH})_3\right]=27 \times 10^{-18} \text { at } 298 \mathrm{~K} .\)
Given: \(\mathrm{K}_{\mathrm{sp}}\left[\mathrm{~A}(\mathrm{OH})_2\right]=9 \times 10^{-10} \text {}\)
\(\mathrm{K}_{\mathrm{sp}}\left[\mathrm{~B}(\mathrm{OH})_3\right]=27 \times 10^{-18} \text { at } 298 \mathrm{~K} .\)
| 1. | B(OH)3 will precipitate before A(OH)2 |
| 2. | A(OH)2 and B(OH)3 will precipitate together |
| 3. | A(OH)2 will precipitate before B(OH)3 |
| 4. | Both A(OH)2 and B(OH)3 do not show precipitation with NH4OH |
Subtopic: Solubility Product | Common Ion Effect |
62%
Level 2: 60%+
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The \(\text{K}_\text{sp}\) of \(\text{Mg(OH)}_2\) is \(1\times 10^{-12},0.01~\text{M Mg}^{2+}\) ion will precipitate at the limiting pH equal to: (at \(25^\circ\text{C}\))
1. 9
2. 8
3. 6
4. 4
1. 9
2. 8
3. 6
4. 4
Subtopic: Solubility Product |
76%
Level 2: 60%+
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If the solubility of AgCl in an aqueous solution is 1.434 x 10-3 M then find the value of [-logKsp] where Ksp is the solubility product of AgCl:
1. 3.7
2. 5.7
3. 6.7
4. 7.7
1. 3.7
2. 5.7
3. 6.7
4. 7.7
Subtopic: Solubility Product |
79%
Level 2: 60%+
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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
1. 3
2. 5
3. 6
4. 1
Subtopic: Solubility Product |
84%
Level 1: 80%+
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At \(310~ K,\) the solubility of \(\mathrm{CaF}_2\) in water is \(2.34 \times 10^{-3} \mathrm{~g} / 100 \mathrm{~ml} \text {. }\)
If the solubility product of \(\mathrm{CaF}_2\) is \(x \times 10^{-10}\), Find \(x\) :
[Given molar mass of \(\mathrm{CaF}_2=78 \mathrm{~g} \mathrm{~mol}^{-1}\) ]
1. 3.00
2. 1.08
3. 0.01
4. 7.20
Subtopic: Solubility Product |
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If the solubility product of PbS is 8 × 10–28, then the solubility of PbS in pure water at 298 K is \(x × 10^{–16}\) .
The value of \(x\) is:
\(\text { [Given : } \sqrt{2}=1.41 \text { ] }\)
1. 262
2. 327
3. 395
4. 282
The value of \(x\) is:
\(\text { [Given : } \sqrt{2}=1.41 \text { ] }\)
1. 262
2. 327
3. 395
4. 282
Subtopic: Solubility Product |
88%
Level 1: 80%+
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The solubility of Ca(OH)2 in water is:
[Given: Ksp Ca(OH)2 in water = 5.5 × 10–6]
1. 1.77 × 10–6
2. 1.11 × 10–6
3. 1.11 × 10–2
4. 2.77 × 10–2
Subtopic: Solubility Product |
74%
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The solubility product of Pbl2 is 8.0 × 10–9 . The solubility of lead iodide in 0.1 molar solution of lead nitrate is x × 10–6 mol/L. The value of x is:
(Rounded off to the nearest integer)
| 1. | 154 | 2. | 423 |
| 3. | 282 | 4. | 141 |
Subtopic: Solubility Product |
66%
Level 2: 60%+
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