Q. No.The solubility product for a salt of type AB is . The molarity of its standard solution will be:
1.
2.
3.
4.
1.
The salt solution that is basic in nature is:
1. Ammonium chloride.
2. Ammonium sulphate.
3. Ammonium nitrate.
4. Sodium acetate.
The pKb of dimethylamine and pka of acetic acid are 3.27 and 4.77 respectively at T (K).
The correct option for the pH of dimethylammonium acetate solution is:
| 1. | 7.75 | 2. | 6.25 |
| 3. | 8.50 | 4. | 5.50 |
Given that the ionic product of is 2 x .
The solubility of in 0.1 M NaOH is ;
1. 2 x M
2. 1 x M
3. 1 x M
4. 2 x M
[Given pKa of CH3COOH = 4.57]
| 1. | 2.57 | 2. | 5.57 |
| 3. | 3.57 | 4. | 4.57 |
\(3 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{O}_{3}(\mathrm{g}) \)
For the above reaction at 298 K, \(K_c\) is found to be \(3.0 \times 10^{-59} \). If the concentration of \(O_2\) at equilibrium is 0.040 M, then the concentration of \(O_3 \) in M is:
1. \(1.2 \times 10^{21} \)
2. \(4.38 \times 10^{-32} \)
3. \(1.9 \times 10^{-63} \)
4. \(2.4 \times 10^{31} \)
The solubility product of \(BaSO_4\) in water is \(1.5 \times 10^{-9} \). The molar solubility of \(BaSO_4\) in 0.1 M solution of Ba(NO3)2 in-
1. \(2.0 \times 10^{-8} M\)
2. \(0.5 \times 10^{-8} M\)
3. \(1.5 \times 10^{-8} M\)
4. \(1.0 \times 10^{-8} M\)
Consider the following reaction taking place in 1L capacity container at 300 K.
\(A +B \rightleftharpoons C+D \)
If one mole each of A and B are present initially and at equilibrium 0.7 mol of C is formed, then the equilibrium constant \((K_c) \) for the reaction is
| 1. | 9.7 | 2. | 1.2 |
| 3. | 6.2 | 4. | 5.4 |
| 1. | 3 | 2. | 2 |
| 3. | 4 | 4. | 1 |
\(CO_{2}(g)\,+\,C(s)\rightarrow \,2CO(g)\)
The value of Kc for the reaction at the same temperature is:
(Given - R = 0.083 L bar K-1 mol-1)
| 1. | 0.36 | 2. | 3.6 × 10-2 |
| 3. | 3.6 × 10-3 | 4. | 3.6 |
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