An aqueous solution of 2 % (w\w) non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. Molar mass of the solute would be:
1. | 23.69 g mol–1 | 2. | 41.35 g mol–1 |
3. | 56.23 g mol–1 | 4. | 22.76 g mol–1 |
Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. The vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane would be:
1. 43.45 kPa
2. 78.96 Pa
3. 73.43 kPa
4. 65.72 Pa
The mass of a non-volatile solute (molar mass 40 g mol–1) that should be dissolved in 114 g octane to reduce its vapour pressure to 80 % would be:
1. 6 g
2. 7 g
3. 8 g
4. 10 g
A 5 % solution (by mass) of cane sugar in water has freezing point of 271 K. Freezing point of 5 % glucose in water would be: (freezing point of pure water is 273.15 K)
1. 279.24 K
2. –269.06 K
3. 275.42 oC
4. 269.06 K
At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. The concentration at which osmotic pressure of the solution becomes 1.52 bar at the same temperature, would be:
1. | 0.06 M | 2. | 0.12 M |
3. | 0.08 M | 4. | 0.36 M |
The correct sequence of order of increasing solubility in the n-octane is:
1. Cyclohexane < CH3CN < CH3OH < KCl
2. KCl < CH3OH < CH3CN < Cyclohexane
3. KCl < CH3OH > CH3CN < Cyclohexane
4. None of the above
If the density of lake water is 1.25 g mL–1 and contains 92 g of Na+ ions per kg of water, then the molality of Na+ ions will be:
1. | 3.24 molal | 2. | 4 molal |
3. | 5 molal | 4. | 3.5 molal |
Benzene and toluene form an ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively.
The mole fraction of benzene in vapour phase, if 80 g of benzene is mixed with 100 g of toluene, would be:
1. 0.41
2. 0.68
3. 0.72
4. 0.59
The partial pressure of ethane over a solution containing 6.56 × 10–2 g of ethane is 1 bar. If the solution contains 5.00 × 10–2 g of ethane, the partial pressure of the gas will be:
1. 0.66 bar
2. 0.96 bar
3. 0.76 bar
4. 0.19 bar
The positive deviations from Raoult’s law mean the vapour pressure is:
1. Higher than expected.
2. Lower than expected.
3. As expected.
4. None of the above