Skip to content
Question

Q.Which of the following colligative property is used to find the molar mass of proteins? (A) Osmotic pressure (B) Elevation in boiling point (C) Depression in freezing point (D) Relative lowering of vapour pressure

CBSECBSE Class XII Board 2023MCQ· 1mImportance★★★★★
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Osmotic pressure is the only colligative property sensitive enough to measure the very small concentrations typical of protein solutions, making it the method of choice for determining the molar mass of macromolecules like proteins.

Why osmotic pressure wins for proteins

Colligative properties depend only on the number of solute particles, not their identity. For a given mass of solute, the magnitude of the effect is inversely proportional to the molar mass — smaller molar mass means more particles, hence a larger effect. Proteins have enormous molar masses (tens of thousands to millions of g/mol), so even a reasonable mass of protein dissolved gives a very small number of moles. That means the changes in boiling point, freezing point, or vapour pressure are tiny — often too small to measure accurately with ordinary instruments.

Osmotic pressure, however, is different. It is directly proportional to the molar concentration at a given temperature, and the proportionality constant (RTRT) is large. For dilute solutions, osmotic pressure can be measured with high precision using a simple manometer or a more sensitive osmometer. This makes it the only practical choice among the four options.

The van’t Hoff equation for osmotic pressure:

Π=iMRT\Pi = i M R T

where Π\Pi is osmotic pressure, ii is the van’t Hoff factor (1 for non-electrolytes like most proteins), MM is molarity (mol/L), RR is the gas constant, and TT is absolute temperature.

Step-by-step reasoning

  1. Recall the four colligative properties

    Relative lowering of vapour pressure (ΔP/P0\Delta P/P_0), elevation in boiling point (ΔTb\Delta T_b), depression in freezing point (ΔTf\Delta T_f), and osmotic pressure (Π\Pi). All four depend on the mole fraction or molar concentration of solute.

  2. Understand the scale of the effect for proteins

    Suppose you dissolve 1 g of a protein of molar mass 50,000 g/mol in 100 mL of water. The number of moles is 1/50000=2×10−51/50000 = 2 \times 10^{-5} mol. The molarity is 2×10−42 \times 10^{-4} M.

    For boiling point elevation: ΔTb=Kb⋅m≈0.512×2×10−4≈1×10−4 ∘C\Delta T_b = K_b \cdot m \approx 0.512 \times 2 \times 10^{-4} \approx 1 \times 10^{-4}\,^\circ\text{C} — far too small to measure with a standard thermometer.

    For freezing point depression: ΔTf=Kf⋅m≈1.86×2×10−4≈3.7×10−4 ∘C\Delta T_f = K_f \cdot m \approx 1.86 \times 2 \times 10^{-4} \approx 3.7 \times 10^{-4}\,^\circ\text{C} — also tiny.

    For osmotic pressure: Π=MRT=(2×10−4)×0.0821×298≈0.0049 atm≈3.7 mm Hg\Pi = MRT = (2 \times 10^{-4}) \times 0.0821 \times 298 \approx 0.0049 \text{ atm} \approx 3.7 \text{ mm Hg}. This is easily measurable with a simple column of mercury or water.

  3. Compare the sensitivities

    The key insight: ΔTb\Delta T_b and ΔTf\Delta T_f are proportional to molality, while Π\Pi is proportional to molarity. But the real difference is the magnitude of the constants. KbK_b and KfK_f are small (around 0.5 and 1.86 for water), while RTRT is about 24.5 L·atm/mol at room temperature — roughly 50 times larger than KfK_f and 100 times larger than KbK_b. This makes osmotic pressure the most sensitive colligative property by far.

  4. Eliminate the other options …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.