Q. No.Match the following type of intermolecular interactions present in the pair of compounds given:
(a)
Methanol and Acetone
(i)
Van der Waal’s forces of attraction
(b)
Acetonitrile and Acetone
(ii)
Ion-dipole interaction
(iii)
Dipole-dipole interaction
(a)
(b)
1.
(iii)
(ii)
2.
(ii)
(ii)
3.
(iii)
(iii)
4.
(iii)
(i)
What mass of urea is needed to prepare 2.5 kg of a 0.25 m aqueous solution?
| 1. | 73 g | 2. | 37 g |
| 3. | 48 g | 4. | 24 g |
The solubility of \(H_2S\) in water at STP is 0.195 m. The value of Henry's constant is:
| 1. | 274 atm | 2. | 285 atm |
| 3. | 295 atm | 4. | 278 atm |
The vapour pressure of water in the solution having 50 g of urea dissolved in 850 g of water is:
(Vapor pressure of pure water at 298 K is 23.8 mm Hg)
1. 23.40 mm of Hg
2. 33.46 mm of Hg
3. 12.76 mm of Hg
4. 87.12 mm of Hg
The solubility of gases in liquids generally decreases as temperature increases.
The primary reason for this behavior is:
| 1. | Dissolution of a gas in a liquid is an endothermic process. |
| 2. | Dissolution of a gas in a liquid is an exothermic process. |
| 3. | Gases are highly compressible. |
| 4. |
All of the above statements are correct. |
Which law is applied when deep-sea divers use a breathing mixture of oxygen and less soluble helium to minimise the painful effects caused by the increased dissolution of gases in blood at high pressure?
| 1. | Raoult's law | 2. | Henry's law |
| 3. | Ideal gas Equation | 4. | All of the above |
| Type of solution | Example | ||
| a. | Solid in gas | i. | Aerated water |
| b. | Gas in liquid | ii. | Smoke |
| c. | Liquid in solid | iii. | Solution of hydrogen in palladium |
| d. | Gas in solid | iv. | Amalgams |
| a | b | c | d | |
| 1. | i | iii | iv | ii |
| 2. | ii | i | iv | iii |
| 3. | iii | i | iv | ii |
| 4. | iv | i | ii | iii |
An example of a gas in a solid-type solution is:
1. Solution of hydrogen in palladium
2. Ethanol dissolved in water
3. Camphor vapours in N2 gas
4. Amalgams
| 1. | \(\text X_{\text{mole fraction}}=\frac{\text n_{\text{solute}}}{\text n_{\text{solution}}}\) |
| 2. | \(\text{Molarity}=\frac{\text{amount of solute (g)}}{\text{volume of solution (mL)}}\) |
| 3. | \(\text{Molality}=\frac{\text{Number of mole of solute}}{\text{amount of solvent (kg)}}\) |
| 4. | \(\text{Mass percentage}=\frac{\text{mass of the component in the solution}}{\text{Total mass of the solution}}\times100 \) |
The density of 68 % nitric acid by mass in an aqueous solution is 1.504 g mL–1. The molarity of the acid solution would be:
1. 15.24 M
2. 16.23 M
3. 14.52 M
4. 13.45 M
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