A 6% solution of glucose (molar mass = 180 g mol$^{-1}$) is isotonic with 2·5% solution of an unknown organic substance. Calculate the molecular weight of the unknown organic substance.

A 6% solution of glucose (molar mass = 180 g mol⁻¹) is isotonic with 2·5% solution of an unknown organic subst

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CBSECBSE Class XII Board 2024Subjective· 2mImportance★★★★★

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Concept understanding — Osmotic Pressure Molar Mass

Osmotic Pressure and Molar Mass: From Intuition to Formula

Imagine you have a glass of pure water, and you carefully place a tea bag into it. After a while, the water turns brown. The tea molecules have moved from the bag into the water. That's simple diffusion. But now imagine a different setup: you have a U-shaped tube with a special membrane at the bottom that only lets water molecules pass through — not larger molecules like sugar. On one side you put pure water, on the other side you put a sugar solution. What happens?

Water will spontaneously move from the pure water side into the sugar solution side, pushing the liquid level higher on the sugar side. That rising column of liquid is a direct physical effect — it's osmotic pressure trying to equalise concentrations. The taller the column gets, the more hydrostatic pressure it exerts back. Eventually, that back-pressure exactly balances the "pull" of the sugar, and the system stops.

That balancing pressure — the pressure you would need to apply to the solution side to prevent the water from moving — is the osmotic pressure ( $\Pi$ ).

The Intuition Behind Molar Mass from Osmotic Pressure

Here's the key insight: the osmotic pressure depends only on the number of solute particles in a given volume of solution, not on what those particles are. A big protein molecule and a tiny sugar molecule, if present in the same number per litre, produce the same osmotic pressure.

This is incredibly useful. If you dissolve an unknown substance (say, a polymer or a protein) in water and measure the osmotic pressure, you can work backwards to find how many moles of it are present. And if you know the mass you dissolved, you can calculate the molar mass:

\text{Molar mass} = \frac{\text{mass of solute (g)}}{\text{number of moles}}

So osmotic pressure becomes a direct window into the molecular weight of substances that are too large or too fragile to vaporise (like proteins, polymers, or enzymes).

The Precise Statement

For dilute solutions, osmotic pressure follows a law that looks exactly like the ideal gas law:

\Pi V = nRT

This is the van't Hoff equation for osmotic pressure. It tells you that osmotic pressure is directly proportional to the molar concentration of the solute:

\Pi = \frac{n}{V}RT = cRT

where $c$ is the molar concentration (mol/L).

From Osmotic Pressure to Molar Mass

If you dissolve a known mass $w$ (in grams) of an unknown substance in a volume $V$ of solvent, and measure the osmotic pressure $\Pi$ at temperature $T$ , you can find the molar mass $M$ as follows:

From $\Pi V = nRT$ , we get $n = \frac{\Pi V}{RT}$
But $n = \frac{w}{M}$ (mass divided by molar mass)
Equating: $\frac{w}{M} = \frac{\Pi V}{RT}$
Rearranging:

M = \frac{wRT}{\Pi V}

This is the working formula. Every quantity on the right is measurable in the lab.

Why This Method is Special

Osmotic pressure measurements are extraordinarily sensitive. For a substance with a very large molar mass (say, 100,000 g/mol), the freezing point depression or boiling point elevation would be too tiny to measure accurately. But osmotic pressure can still give a measurable reading because it's a colligative property that depends only on particle count, and the effect is large even at low concentrations.

Important

Osmotic pressure is the most sensitive colligative property for determining molar masses of macromolecules. It can detect concentrations as low as $10^{-4}$ M, which is 100–1000 times more sensitive than freezing point depression.

A Worked Example

Problem: 0.50 g of a protein is dissolved in enough water to make 100 mL of solution at 25°C. The osmotic pressure is measured as 0.012 atm. Find the molar mass of the protein.

Solution:

Given: …

Why this formula?

Great — let’s build a clear, concept-first understanding of Osmotic Pressure and its link to Molar Mass.

1. What is Osmotic Pressure?

Osmotic pressure ( $\Pi$ ) is the minimum pressure that must be applied to a solution to prevent the net flow of solvent into it through a semipermeable membrane.

Think of it as the “push” needed to stop the solvent from diluting the solution.

2. The Key Formula

The central equation is:

\Pi = i \, C \, R \, T

Where:

$\Pi$ = osmotic pressure (atm or Pa)
$i$ = van’t Hoff factor (number of particles per formula unit)
$C$ = molar concentration (mol/L or mol/m³)
$R$ = universal gas constant
$T$ = absolute temperature (K)

For non-electrolytes (like glucose, urea), $i = 1$ , so:

\Pi = C R T

3. Why does this formula hold? — The Reasoning

Step 1: Analogy to Ideal Gas Law

The van’t Hoff equation for osmotic pressure is structurally identical to the ideal gas law:

P V = n R T \quad \Rightarrow \quad P = \frac{n}{V} R T = C R T

Why? Because solute particles in a dilute solution behave like gas molecules — they are far apart, move randomly, and exert a “pressure” on the membrane.

In a gas: particles hit the container walls → pressure.
In a solution: solute particles cannot cross the membrane, but they collide with it → osmotic pressure.

So, the formula $\Pi = C R T$ is not a coincidence — it’s a direct analogy.

Step 2: The van’t Hoff Factor $i$

For electrolytes (e.g., NaCl), one formula unit dissociates into multiple ions:

NaCl → Na⁺ + Cl⁻ → $i = 2$
CaCl₂ → Ca²⁺ + 2Cl⁻ → $i = 3$

Each ion acts as an independent particle, so the effective concentration increases by factor $i$ :

\Pi = i \, C \, R \, T

Step 3: Linking to Molar Mass

We usually know mass of solute ( $w$ ) and volume of solution ( $V$ ). Molar concentration is:

C = \frac{n}{V} = \frac{w / M}{V}

where $M$ = molar mass (g/mol).

Substitute into the osmotic pressure equation:

\Pi = i \cdot \frac{w}{M V} \cdot R T

Rearrange to solve for molar mass:

\boxed{M = \frac{i \, w \, R \, T}{\Pi \, V}}

This is the key formula used in experiments to find molar mass from osmotic pressure.

4. Why is this method special? …

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Two solutions are isotonic when they exert equal osmotic pressure, which happens when they have the same molar concentration. Equating the molarities of the glucose and unknown solutions gives a molecular weight of 75 g mol⁻¹ for the unknown substance.

Why isotonic solutions share the same molar concentration

Osmotic pressure $\pi$ is a colligative property — it depends only on the number of solute particles, not their identity. For dilute solutions, van 't Hoff's equation tells us:

\pi = CRT

where $C$ is the molar concentration (mol L⁻¹), $R$ is the gas constant, and $T$ is temperature.

When two solutions are isotonic, they exert identical osmotic pressure across a semipermeable membrane. Since $R$ and $T$ are the same for both solutions at equilibrium, the equation immediately reveals that their molar concentrations must be equal:

C_{\text{glucose}} = C_{\text{unknown}}

This equality is the key. We can calculate each concentration from the percentage composition and molar mass, then solve for the unknown molar mass.

Step-by-step calculation

1. Interpret the percentage concentrations

A 6% solution of glucose means 6 g of glucose in 100 mL of solution. Similarly, a 2.5% solution of the unknown means 2.5 g of unknown in 100 mL of solution. We'll work with these 100 mL volumes throughout.

2. Calculate the molar concentration of glucose

The number of moles of glucose in 100 mL is:

n_{\text{glucose}} = \frac{6 \text{ g}}{180 \text{ g mol}^{-1}} = \frac{1}{30} \text{ mol}

Since this is dissolved in 100 mL = 0.1 L, the molarity is:

C_{\text{glucose}} = \frac{1/30 \text{ mol}}{0.1 \text{ L}} = \frac{1}{3} \text{ mol L}^{-1}

3. Set up the equation for the unknown substance

Let the molar mass of the unknown be $M$ g mol⁻¹. The number of moles in 100 mL is:

n_{\text{unknown}} = \frac{2.5}{M} \text{ mol}

Its molarity is: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2024Set 56/3/11 markMCQ
$mol^{-1}$
›Reveal solutionSolution
Isotonic solutions have the same osmotic pressure, which for dilute non‑electrolytes means equal molar concentrations. Equating the molarities of the 1% X solution and the 6% sucrose solution gives the molar mass of X as 57 g mol⁻¹.

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.

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.

Set the molarities equal because the solutions are isotonic.

$\frac{10}{M_X} = \frac{60}{342}$

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

$3420 = 60\,M_X$
$M_X = \frac{3420}{60} = 57$ …
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CBSE 2023Set 56/1/11 markMCQ

Q. The colligative property used for the determination of molar mass of polymers and proteins is : (A) Osmotic pressure (B) Depression in freezing point (C) Relative lowering in vapour pressure (D) Elevation in boiling point

›Reveal solution

Osmotic pressure is the only colligative property with a magnitude large enough to measure accurately for high-molar-mass polymers and proteins in dilute solution. The answer is (A).

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:

Property	Proportionality constant	Effect for $m \approx 0.0001$
Depression in freezing point	$K_f \approx 1.86 \, \text{K kg mol}^{-1}$ (water)	$\Delta T_f \approx 0.0002 \, \text{K}$
Elevation in boiling point	$K_b \approx 0.52 \, \text{K kg mol}^{-1}$ (water)	$\Delta T_b \approx 0.00005 \, \text{K}$
Relative lowering of vapour pressure	$\frac{\Delta p}{p^0} = \chi_{\text{solute}}$	$\approx 0.000002$
Osmotic pressure	$\pi = CRT$	$\pi \approx 2.5 \, \text{kPa}$ at $298 \, \text{K}$

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.

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.
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.
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).

Tip

Osmotic pressure is the only colligative property that remains experimentally accessible at the low particle concentrations characteristic of polymer and protein solutions. …

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CBSE 2023Set 56/2/11 markMCQ
$Q. Given below are two statements labelled as Assertion (A) and Reason (R). Select the most appropriate answer from the options given below : Assertion (A) : Osmotic pressure is a colligative property. Reason (R) : Osmotic pressure is proportional to the molality. (A) Both (A) and (R) are true and (R) is the correct explanation of (A). (B) Both (A) and (R) are true, but (R) is not the correct explanation of (A). (C) (A) is true, but (R) is false. (D) (A) is false, but (R) is true.$
›Reveal solutionSolution
Osmotic pressure is a colligative property because it depends only on the number of solute particles, not their identity. The Reason says it is proportional to molality — this is true only for ideal dilute solutions, but the statement is incomplete and misleading in the context of the Assertion. The correct answer is (B).

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.

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.

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.

Watch out

A common mistake is to treat molality and molarity as interchangeable. They are not. Osmotic pressure is directly proportional to molarity (moles per litre of solution), not molality. The Reason’s wording is therefore inaccurate in a strict sense.

Connecting the two: …
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CBSE 2023Set 56/3/11 markMCQ
$Q. Which of the following colligative property is used to find the molar mass of proteins? (A) Osmotic pressure (B) Elevation in boiling point (C) Depression in freezing point (D) Relative lowering of vapour pressure$
›Reveal solutionSolution
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 ( $RT$ ) 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.
$\Pi = i M R T$
Step-by-step reasoning

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.

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.

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.

Eliminate the other options …
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CBSE 2020Set 56/3/11 markMCQ
$Q. Assertion (A) : Osmotic pressure is a colligative property. Reason (R) : Osmotic pressure is directly proportional to molarity. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.$
›Reveal solutionSolution

Osmotic pressure is indeed a colligative property (depends on particle number, not identity), and it is directly proportional to molarity — but the proportionality to molarity doesn't explain why it's colligative; both facts are true yet logically independent.

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
$\pi = i \, C \, R \, T$
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. …
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Q.A 6% solution of glucose (molar mass = 180 g mol $^{-1}$ ) is isotonic with 2·5% solution of an unknown organic substance. Calculate the molecular weight of the unknown organic substance.

Concept understanding — Osmotic Pressure Molar Mass

Osmotic Pressure and Molar Mass: From Intuition to Formula

The Intuition Behind Molar Mass from Osmotic Pressure

The Precise Statement

From Osmotic Pressure to Molar Mass

Why This Method is Special

A Worked Example

Why this formula?

1. What is Osmotic Pressure?

2. The Key Formula

3. Why does this formula hold? — The Reasoning

Step 1: Analogy to Ideal Gas Law

Step 2: The van’t Hoff Factor $i$

Step 3: Linking to Molar Mass

4. Why is this method special? …

Concept: Isotonic Solutions and Osmotic Pressure

Why isotonic solutions share the same molar concentration

Step-by-step calculation

Why osmotic pressure works for macromolecules

Why the magnitude difference?

Why osmotic pressure wins for proteins

Step-by-step reasoning

Why osmotic pressure is colligative

The van 't Hoff equation

Evaluating the assertion and reason

Does (R) explain (A)?

Q.A 6% solution of glucose (molar mass = 180 g mol−1^{-1}−1) is isotonic with 2·5% solution of an unknown organic substance. Calculate the molecular weight of the unknown organic substance.

Why isotonic solutions share the same molar concentration

Step-by-step calculation

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Q.A 6% solution of glucose (molar mass = 180 g mol $^{-1}$ ) is isotonic with 2·5% solution of an unknown organic substance. Calculate the molecular weight of the unknown organic substance.

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