A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. The mass percentage of the solvent in resulting solution is:
1. 66.43 %
2. 36.54 %
3. 12.45 %
4. 33.97 %
The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. The composition of the liquid mixture if total vapour pressure is 600 mm Hg is:
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 amount of sucrose is to be added to 500 g of water such that it boils at 100°C is:
( Given: The boiling point of water at 750 mm Hg is 99.63°C ; Kb = 0.52 K kg mol-1)
1. 145.76 g
2. 213.54 g
3. 121.67 g
4. 195.36 g
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 |
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. | \(\text{Mass percentage}=\frac{\text{mass of the component in the solution}}{\text{Total mass of the solution}}\times100 \) |
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