Q. No.The solubility of H2S 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 solubility of a gas in a liquid decreases with an increase in temperature because:
| 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. |
To minimize the painful effects accompanying deep sea diving, oxygen diluted with less soluble helium gas is used as breathing gas by the divers. This is an example of the application of:
| 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 |
A solution of glucose (M.W = 180 g mol–1) in water is 10 % w/w. The mole fraction of each component in the solution is:
| 1. | Mole fraction of glucose - 0.44; mole fraction of water - 0.56 |
| 2. | Mole fraction of glucose - 0.056; mole fraction of water - 0.944 |
| 3. | Mole fraction of glucose - 0.011; mole fraction of water - 0.99 |
| 4. | Mole fraction of glucose - 0.36; mole fraction of water - 0.64 |
The type of molecular interactions in a solution of alcohol and water is:
| 1. | Electrostatic interaction | 2. | London dispersion forces |
| 3. | Hydrogen bonding | 4. | Ion-dipole interaction |
The graph representing Henry's law is given below:
The value of the slope will be:
| 1. | \(4\times10^{-4}~\mathrm{torr}\) | 2. | \(5\times10^{-4}~\mathrm{torr}\) |
| 3. | \(4\times10^{4}~\mathrm{torr}\) | 4. | \(5\times10^{4}~\mathrm{torr}\) |
Henry’s law constant for the solution of methane in benzene at 298 K is 4.27 × 105 mm Hg. The mole fraction of methane in benzene at 298 K under 760 mm Hg will be:
1. 1.85 × 10–5
2. 192 × 10–4
3. 178 × 10–5
4. 18.7 × 10–5
The amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol is:
| 1. | 4.57 g | 2. | 3.57 g |
| 3. | 1.57 g | 4. | 12.57 g |
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