Q. No.If 22 g of benzene is dissolved in 122 g of CCl4, the mass percentage of CCl4 and benzene respectively are-
1.
2.
3.
4. 91.28 %, 8.72 %
The type of inter molecular interactions present in
| (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) |
To make 2.5 kg of 0.25 m aqueous solution, the mass of urea required is-
| 1. | 73 g | 2. | 37 g |
| 3. | 48 g | 4. | 24 g |
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 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 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 minimise 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 |
The example of 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. | \(\text{Mass percentage}=\frac{\text{mass of the component in the solution}}{\text{Total mass of the solution}}\times100 \) |
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