Q. No.| 1. | \(s=\left(\frac{K_{s p}}{106}\right)^{1 / 4}~ \) | 2. | \(s=\left(\frac{K_{s p}}{108}\right)^{1 / 5} \) |
| 3. | \(s=\left(\frac{K_{s p}}{81}\right)^{1 / 5} \) | 4. | \(s=\left(\frac{K_{s p}}{48}\right)^{1 / 5} \) |
The solubility product for a salt of the type AB is 4 × . What is the molarity of its standard solution?
1.
2.
3.
4.
1. 4X3
2. 108X5
3. 27X4
4. 9X
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| 1. | \( \mathrm{S}=\left(\frac{\mathrm{K}_{s p}}{144}\right)^{1 / 6} \) | 2. | \( \mathrm{~S}=\left(\frac{\mathrm{K}_{s p}}{6912}\right)^{1 / 7}\) |
| 3. | \( \mathrm{~S}=\left(\frac{\mathrm{K}_{s p}}{929}\right)^{1 / 9} \) | 4. | \( \mathrm{~S}=\left(\frac{\mathrm{K}_{s p}}{216}\right)^{1 / 7} \) |
1. \(1.3 \times 10^{-10} \mathrm{M}\)
2. \(4.5 \times 10^{-6} \mathrm{M}\)
3. \(1.3 \times 10^{-5} \mathrm{M}\)
4. \(1.3 \times 10^{-7} \mathrm{M}\)
| Assertion (A): | The solubility of AgCl in \(NH_3 (aq.) \) is higher than that in pure water. |
| Reason (R): | When AgCl dissolves in \(NH_3 (aq.) \), it forms a complex ion \(\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}\), which shifts the solubility equilibrium of AgCl in the forward direction. |
| 1. | Both (A) and (R) are True and (R) is the correct explanation of (A). |
| 2. | Both (A) and (R) are True but (R) is not the correct explanation of (A). |
| 3. | (A) is True but (R) is False. |
| 4. | (A) is False but (R) is True. |
The solubility product of a sparingly soluble salt AX2 is 3.2 ×10–11. Its solubility (in moles/litre) is:
| 1. | 3.1×10–4 | 2. | 2 × 10–4 |
| 3. | 4 × 10–4 | 4. | 5.6 × 10–6 |
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Calculate the solubility of AgI in a 10⁻⁴ N KI solution at 25°C, given that the solubility product constant (Kₛₚ) of AgI is 1.0 × 10⁻¹⁶ mol² L⁻².
1.
2.
3.
4.
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The solubility of AgCl (s) with solubility product 1.6×10–10 in 0.1 M NaCl solution would be?
| 1. | 1.26 × 10–5 M | 2. | 1.6 × 10–9 M |
| 3. | 1.6 × 10–11 M | 4. | zero |
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(Given: Ksp for BaSO4 = 1.1 × 10-10)
| 1. | 1.1 × 10–13 M | 2. | 1.1 × 10–7 M |
| 3. | 5.5 × 10–7 M | 4. | 5.5 × 10–8 M |
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