The vapour pressure of a solvent decreased by 10 mm of Hg when a non-volatile solute was added to the solvent. The mole fraction of the solute in solution is 0.2. What would be the mole fraction of the solvent if the decrease in vapour pressure is 20 mm of Hg:
1. 0.2
2. 0.4
3. 0.6
4  0.8

Camphor is often used in molecular mass determination because:

1. It has a very high cryoscopic constant

2. It is volatile

3. It is a solvent for organic substances

4. It is readily available 

From the colligative properties of a solution, which one is the best method for the determination of molecular weight of proteins & polymers:

1. Osmotic pressure

2. Lowering of vapour pressure 

3. Lowering of freezing point

4. Elevation in boiling point

A solution contains a non-volatile solute of molecular mass M₂. The molecular mass of solute in terms of osmotic pressure is:
where:

m₂ → Mass of solute

V → Volume of solution

\(\pi\) → Osmotic pressure

1. ${\mathrm{M}}_2=\left(\frac{{\mathrm{m}}_2}{\mathrm{\pi }}\right)\mathrm{VRT}$

2. ${\mathrm{M}}_2=\left(\frac{{\mathrm{m}}_2}{\mathrm{V}}\right)\frac{\mathrm{RT}}{\mathrm{\pi }}$

3. ${\mathrm{M}}_2=\left(\frac{{\mathrm{m}}_2}{\mathrm{V}}\right)\mathrm{\pi RT}$

4. ${\mathrm{M}}_2=\left(\frac{{\mathrm{m}}_2}{\mathrm{V}}\right)\frac{\mathrm{\pi }}{\mathrm{RT}}$

The ideal solution indicates:

Beans get cooked earlier in a pressure cooker, because:

If 22 g of benzene is dissolved in 122 g of carbon tetrachloride then the mass percentage of carbon tetrachloride (CCl₄) is-

1. 82.72%

2. 83.72%

3. 84.72%

4. 85.72%

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)