At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmot

Exercises · 1.22

Q. At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?

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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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Osmotic pressure is directly proportional to molar concentration at constant temperature ( $\pi = CRT$ ), so the new concentration is found by scaling the known concentration by the ratio of the two osmotic pressures. The concentration of the second solution is $\boxed{C_2 \approx 0.0610\ \text{mol L}^{-1}}$ .

Concept: Osmotic Pressure and the van't Hoff Equation

Osmotic pressure ( $\pi$ ) is the pressure that must be applied to a solution to stop the net flow of solvent across a semipermeable membrane from a pure solvent (or a more dilute solution) into it. For a dilute solution, osmotic pressure obeys the van't Hoff equation, which has the same form as the ideal gas equation:

\pi = CRT

where $C$ is the molar concentration of the solution (mol L $^{-1}$ ), $R$ is the universal gas constant, and $T$ is the absolute temperature.

Since both solutions in this problem are at the same temperature ( $T = 300\ \text{K}$ ) and $R$ is a universal constant, $\pi$ depends only on $C$ — the two quantities are directly proportional:

\pi \propto C \quad \Rightarrow \quad \frac{\pi_1}{C_1} = \frac{\pi_2}{C_2} = RT

This proportionality means we don't even need the numerical value of $R$ : we can find the unknown concentration directly by comparing the two states.

Step 1: Find the concentration of the first solution

The first solution contains 36 g of glucose (molar mass $= 180\ \text{g mol}^{-1}$ ) dissolved to make 1 litre of solution:

C_1 = \frac{\text{mass}}{\text{molar mass} \times \text{volume (L)}} = \frac{36\ \text{g}}{180\ \text{g mol}^{-1} \times 1\ \text{L}} = 0.2\ \text{mol L}^{-1}

This solution has osmotic pressure $\pi_1 = 4.98\ \text{bar}$ at $T = 300\ \text{K}$ .

Step 2: Set up the proportionality for the second solution …

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Here are the common mistakes students make when solving osmotic pressure problems like this one, along with how to avoid each.

Mistake 1: Forgetting that Osmotic Pressure is a Colligative Property

The error: Students often try to use the given data (36 g, 4.98 bar) to find the molar mass of glucose, then use that molar mass to find the new concentration. This is unnecessary and wastes time.
Why it’s wrong: Osmotic pressure depends only on the number of particles (moles) of solute, not on the identity or molar mass of the solute. The first data point is just a calibration — it tells you the relationship between concentration and osmotic pressure.
How to avoid: Recognise that for a non-electrolyte like glucose, $\Pi = cRT$ . Since $R$ and $T$ are constant, $\Pi$ is directly proportional to $c$ . Use the ratio method.

Mistake 2: Using the Wrong Formula or Units

The error: Plugging numbers into $\Pi = iCRT$ without checking units. For example, using volume in mL instead of litres, or forgetting that $R$ must match the units of pressure (bar, atm, etc.).
Why it’s wrong: Osmotic pressure in bar requires $R = 0.08314 \ \text{L bar mol}^{-1} \text{K}^{-1}$ . Using $R = 0.0821$ (for atm) gives a wrong answer.
How to avoid: Always write the formula first, then check units. For this problem, since both pressures are in bar and temperature is same, you can skip $R$ entirely by using the ratio.

Mistake 3: Calculating Molar Mass Unnecessarily

The error: Computing molar mass of glucose from the first data point:

M = \frac{wRT}{\Pi V} = \frac{36 \times 0.08314 \times 300}{4.98 \times 1} \approx 180 \ \text{g/mol}

Then using this to find moles in the second case.

Why it’s wrong: It works here (glucose molar mass is indeed 180 g/mol), but it’s a waste of time and can introduce rounding errors. In an exam, this extra step increases the chance of arithmetic mistakes.
How to avoid: Use the direct proportionality:

\frac{\Pi_1}{c_1} = \frac{\Pi_2}{c_2}

where $c_1 = \frac{36}{180} = 0.2 \ \text{mol/L}$ . Then:

c_2 = \frac{\Pi_2}{\Pi_1} \times c_1 = \frac{1.52}{4.98} \times 0.2

Mistake 4: Misinterpreting “Concentration”

The error: Giving the answer in g/L instead of mol/L (molarity), or vice versa.
Why it’s wrong: The question asks for “concentration” — in osmotic pressure problems, this almost always means molar concentration (mol/L). If they wanted mass concentration, they’d specify.
How to avoid: Read the question carefully. If it says “concentration” without qualification, assume molarity (M). If needed, convert at the end.

Mistake 5: Rounding Too Early …

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

GUJCET 2025Set 031 markMCQ
$Q.______ solution is hypertonic with reference to fluid inside the blood cell. (A) 0.8% W/V NaCl (B) 0.6% W/V NaCl (C) 0.9% W/V NaCl (D) 1.2% W/V NaCl$
›Reveal solutionSolution
[!TLDR]

0.9% NaCl is isotonic with blood, so 1.2% NaCl is hypertonic.

Concept

A hypertonic solution has a higher solute concentration (higher osmotic pressure) than the cell fluid, drawing water out of the cell.

Solution

The fluid inside a blood cell is isotonic with 0.9% W/V NaCl (normal saline). Comparing the options:

0.6% and 0.8% NaCl are more dilute than 0.9%, so they are hypotonic. …
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GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
$Q. 400 cm3 of an aqueous solution of a protein contains 1.26 g of the protein. The osmotic pressure of such a solution at 300 K is found to be 2.57 x 10^-3 bar. The molar mass of protein is _____ g mol-1.$
(a) 30519
(b) 51538
(c) 61038
(d) 40519
›Reveal solutionSolution

Using the osmotic pressure equation pi = (w/M)(RT/V), rearranged to solve for molar mass M, gives the protein's molar mass from the given data.

Osmotic pressure equation: pi = (n/V)RT = (w/(M x V)) x RT

Rearranged for molar mass: M = (w x R x T) / (pi x V)

Given: w = 1.26 g, V = 400 cm3 = 0.400 L, pi = 2.57 x 10^-3 bar, T = 300 K, R = 0.083 L bar K-1 mol-1.

M = (1.26 x 0.083 x 300) / (2.57 x 10^-3 x 0.400)

Numerator = 1.26 x 0.083 x 300 = 31.374 …

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GUJCET 2023Set 091 markMCQ
$\pi$
›Reveal solutionSolution
[!TLDR]

With $i=2$ for NaCl, $\pi = iCRT = 0.04\,RT$ .

Concept

For a dilute solution, osmotic pressure $\pi = iCRT$ , where $C$ is molar concentration and $i$ is the van't Hoff factor (number of particles produced per formula unit).

Solution …
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GUJCET 2015Set C1 markMCQ
$Q. Which colligative property is more useful to determine the molecular weight of the substances like proteins and polymers? (A) Elevation in boiling point (B) Lowering of vapour pressure (C) Depression of freezing point (D) Osmotic pressure$
›Reveal solutionSolution
[!TLDR]

Osmotic pressure is the colligative property best suited to determine molar masses of proteins and polymers.

Concept

Colligative properties (NCERT/GSEB solutions) depend on the number of solute particles. Molar mass $=\dfrac{w_2 RT}{\Pi V}$ from osmotic pressure. For macromolecules, molality is tiny, so $\Delta T_b$ , $\Delta T_f$ and vapour-pressure lowering are too small to measure accurately, but osmotic pressure ( $\Pi$ ) is appreciable and measurable at room temperature.

Solution …
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Q.At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?

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: Osmotic Pressure and the van't Hoff Equation

Step 1: Find the concentration of the first solution

Step 2: Set up the proportionality for the second solution …

Method: Direct Proportionality (Osmotic Pressure–Concentration Relation)

Why This Method Works

Steps

Mistake 1: Forgetting that Osmotic Pressure is a Colligative Property

Mistake 2: Using the Wrong Formula or Units

Mistake 3: Calculating Molar Mass Unnecessarily

Mistake 4: Misinterpreting “Concentration”

Mistake 5: Rounding Too Early …

Concept

Solution

Concept

Solution …

Concept

Solution …

Q.At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?

Concept: Osmotic Pressure and the van't Hoff Equation

Step 1: Find the concentration of the first solution

Step 2: Set up the proportionality for the second solution …

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