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
The vapour pressure of 1 molal solution of a non-volatile solute in water at 300 K would be:
(The vapour pressure of water at 300 K = 12.3 kPa)
1. | 21.08 kPa | 2. | 12.08 kPa |
3. | 33.08 kPa | 4. | 4.08 kPa |
A solution containing 30 g of non-volatile solute in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. The vapour pressure of water at 298 K will be -
1) 1.53 kPa
2) 2.53 kPa
3) 3.53 kPa
4) 4.53 kPa
Two elements A and B form compounds having formula AB2 and AB4. When dissolved in 20 g of benzene (C6H6), 1 g of AB2 lowers the freezing point by 2.3 K whereas 1.0 g of AB4 lowers it by 1.3 K. The atomic masses of A and B are respectively:
(Kf for benzene is 5.1 K kg mol-1)
1. 15.59 u and 52.64 u
2. 25.59 u and 42.64 u
3. 13.59 u and 52.64 u
4. 23.59 u and 32.64 u
The type of inter-molecular interactions present in:
(a) | n-Hexane and n-octane | (i) | Van der Waal’s forces of attraction |
(b) | NaClO4 and water | (ii) | Ion-dipole interaction |
(iii) | Dipole-dipole interaction |
(a) | (b) | |
1. | (i) | (ii) |
2. | (ii) | (ii) |
3. | (i) | (iii) |
4. | (iii) | (iii) |
The insoluble compound in water is /are:
1. Phenol
2. Formic acid and toluene
3. Phenol and toluene
4. Toluene and chloroform
The mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when 6.5 g of C9H8O4 is dissolved in 450 g of CH3CN will be:
1. 1.424%
2. 4.424%
3. 5.124%
4. 2.124%
Nalorphene (C19H21NO3), similar to morphine, is used to combat withdrawal symptoms in narcotic users. The dose of nalorphene generally given is 1.5 mg.
Calculate the mass of 1.5 × 10−3m aqueous solution required for the above dose.
1. 13.22 g
2. 3.22 g
3. 11.22 g
4. 9.22 g
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 |