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

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 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

100 g of liquid A (molar mass 140 g mol^–1) was dissolved in 1000 g of liquid B (molar mass 180 g mol^–1). The vapour pressure of pure liquid B was found to be 500 torr. If the total vapour pressure of the solution is 475 torr, the vapour pressure of pure liquid A will be:

The type of molecular interactions in a solution of alcohol and water is: