Q.A solution is prepared by dissolving 5 g of a non-volatile solute in 200 g of water. It has a vapour pressure of 31·84 mm Hg at 300 K. Calculate the molar mass of the solute. (Vapour pressure of pure water at 300 K = 32 mm Hg)

The key idea here is that isotonic solutions exert the same osmotic pressure. For dilute solutions of non‑electrolytes (like sucrose and the unknown solute X), osmotic pressure is given by $\Pi = iCRT$, and since neither solute dissociates, $i = 1$. So $\Pi$ depends only on the molar concentration $C$ (in mol L⁻¹) at a given temperature. If two solutions are isotonic, their molar concentrations must be equal.

The problem gives us percentage concentrations — 1% of X and 6% of sucrose. A “1% solution” means 1 g of solute in 100 mL of solution (or equivalently 10 g per litre). Similarly, 6% sucrose means 6 g per 100 mL, i.e. 60 g per litre. We can convert these mass‑per‑volume concentrations into molarities using the molar mass, and then set them equal.

Let’s work through it step by step.

1. Write the expression for molarity of each solution.

   Molarity $M = \frac{\text{mass of solute per litre (g L⁻¹)}}{\text{molar mass (g mol⁻¹)}}$.

   For sucrose: $M_{\text{sucrose}} = \frac{60}{342}$ mol L⁻¹.

   For X: $M_X = \frac{10}{M_X}$ mol L⁻¹, where $M_X$ is the unknown molar mass in g mol⁻¹.

2. Set the molarities equal because the solutions are isotonic.

3. Solve for $M_X$. Cross‑multiply: $10 \times 342 = 60 \times M_X$

$M_X = \frac{3420}{60} = 57$ …

Why osmotic pressure works for macromolecules

Colligative properties depend on the number of solute particles, not their identity. For a given mass concentration, a high-molar-mass substance produces far fewer particles than a low-molar-mass one. This creates a measurement challenge: the colligative effect becomes vanishingly small.

Consider a $1\%$ solution of a polymer with molar mass $M = 100{,}000 \, \text{g mol}^{-1}$. The molality is roughly $0.0001 \, \text{mol kg}^{-1}$. Now compare the four colligative properties:

The first three produce changes of order $10^{-4}$ to $10^{-6}$, far below the precision of standard thermometers or manometers. Osmotic pressure, however, generates a measurable pressure difference even at very low concentrations.

Why the magnitude difference?

The key lies in the units and the nature of the measurement.

1. Freezing-point depression and boiling-point elevation scale with molality through cryoscopic and ebullioscopic constants that are typically $1$–$2 \, \text{K kg mol}^{-1}$. For macromolecules, $m \sim 10^{-4}$ yields $\Delta T \sim 10^{-4} \, \text{K}$, which requires extraordinarily sensitive thermometry.

2. Vapour-pressure lowering is proportional to mole fraction. For dilute solutions of high-molar-mass solutes, $\chi_{\text{solute}} \sim 10^{-6}$, making the relative change in vapour pressure unmeasurable with ordinary equipment.

3. Osmotic pressure obeys $\pi = CRT$, where $C$ is molar concentration. Even though $C$ is small, the gas constant $R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1}$ and room temperature $T \approx 300 \, \text{K}$ combine to give $RT \approx 2500 \, \text{J mol}^{-1} = 2500 \, \text{Pa L mol}^{-1}$. A concentration of $C = 0.001 \, \text{mol L}^{-1}$ produces $\pi \approx 2.5 \, \text{kPa}$, easily measured with a simple manometer (a column height of about $25 \, \text{cm}$ of water).

Osmotic pressure is one of the four classic colligative properties (along with vapour pressure lowering, boiling point elevation, and freezing point depression). A property is called colligative when its magnitude depends solely on the number of solute particles present in a given amount of solvent, and not on what those particles are. For osmotic pressure, the underlying reason is that the solvent's tendency to move across a semipermeable membrane is governed by its mole fraction — which changes only with the count of solute particles.

Now, the Reason claims that osmotic pressure is proportional to molality. This is true under the ideal dilute solution approximation, where the van’t Hoff equation $\Pi = i M R T$ (with $M$ as molarity) can be approximated using molality for very dilute aqueous solutions. But the Assertion is about why osmotic pressure is colligative — and that reason is fundamentally about particle number, not about proportionality to molality. The two statements are both true in their own right, but the Reason does not explain the Assertion.

Let’s examine each statement carefully.

1. Assertion (A): “Osmotic pressure is a colligative property.”

   This is correct. For a given solvent and temperature, the osmotic pressure $\Pi$ depends only on the concentration of solute particles (ions or molecules), not on their chemical nature. For example, a 0.1 M glucose solution and a 0.1 M urea solution exert the same osmotic pressure (assuming ideal behaviour), because both have the same number of particles per litre.

2. Reason (R): “Osmotic pressure is proportional to the molality.”

   This statement is true only under specific conditions — for ideal, very dilute solutions where molarity ≈ molality. The exact van’t Hoff equation is $\Pi = i M R T$, where $M$ is molarity (moles per litre of solution), not molality (moles per kg of solvent). In dilute aqueous solutions, the numerical difference between molarity and molality is small, so proportionality to molality is approximately true. But strictly speaking, the correct proportionality is to molarity. Hence, the Reason is not universally true — it is an approximation, and in many exam contexts, it is considered false because the precise relationship uses molarity.

3. Connecting the two: …

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 ($RT$) 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.

Step-by-step reasoning

1. Recall the four colligative properties

   Relative lowering of vapour pressure ($\Delta P/P_0$), elevation in boiling point ($\Delta T_b$), depression in freezing point ($\Delta T_f$), and osmotic pressure ($\Pi$). 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 $1/50000 = 2 \times 10^{-5}$ mol. The molarity is $2 \times 10^{-4}$ M.

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

   For freezing point depression: $\Delta T_f = K_f \cdot m \approx 1.86 \times 2 \times 10^{-4} \approx 3.7 \times 10^{-4}\,^\circ\text{C}$ — also tiny.

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

3. Compare the sensitivities

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

4. Eliminate the other options …

Why osmotic pressure is colligative

A colligative property depends only on the number of solute particles in solution, not on their chemical nature. The four classic colligative properties are vapor-pressure lowering, boiling-point elevation, freezing-point depression, and osmotic pressure.

Osmotic pressure arises when a semipermeable membrane separates a solution from pure solvent. Solvent molecules cross the membrane to dilute the solution, creating a hydrostatic pressure difference. The key insight: this pressure depends on how many particles are "blocking" solvent sites on the solution side, regardless of what those particles are. A mole of glucose exerts the same osmotic pressure as a mole of sucrose at the same concentration and temperature.

The van 't Hoff equation

The quantitative relationship is

where $\pi$ is osmotic pressure, $i$ is the van 't Hoff factor (number of particles per formula unit), $C$ is molarity (mol/L), $R$ is the gas constant, and $T$ is absolute temperature.

Because $C$ counts particles per unit volume and $i$ accounts for dissociation, the product $iC$ is the total particle concentration. This confirms osmotic pressure is colligative.

Evaluating the assertion and reason

Assertion (A): Osmotic pressure is a colligative property.

This is true — it depends on particle number, not particle identity.

Reason (R): Osmotic pressure is directly proportional to molarity.

This is also true — the van 't Hoff equation shows $\pi \propto C$ (at constant $T$ and $i$).

Does (R) explain (A)?

Here's the subtle point. Saying "$\pi$ is proportional to molarity" tells us the mathematical form of the relationship. But it doesn't explain why osmotic pressure is colligative. …