The following solutions were prepared by dissolving 10 g of glucose (C₆H₁₂O₆)
in 250 ml of water (P₁), 10 g of urea (CH₄N₂O) in 250 ml of water (P₂) and 10 g
of sucrose (C₁₂H₂₂O₁₁) in 250 ml of water (P₃). The decreasing order of osmotic
the pressure of these solutions is:

Isotonic solutions have the same:

1. Vapour pressure

2. Freezing temperature

3. Osmotic pressure

4. Boiling temperature

200 mL of an aqueous solution contains 1.26 g of protein. The osmotic pressure of this solution at 300 K is found to be 2.57 × 10^–3 bar. The molar mass of protein will be:

(R = 0.083 L bar mol^–1 K^–1):

During osmosis, the flow of water through a semi-permeable membrane is:

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

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

1 % (w/w) solution of a compound is isotonic with 5 % (w/w) sucrose (sugar) solution. The molecular weight of the compound will be: