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)

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 molar mass of the solute will be -

1.  23 g mol⁻¹

2.  34 g mol⁻¹

3.  15 g mol⁻¹

4.  46 g mol⁻¹

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 -

(K_f 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