Determine the osmotic pressure of a solution prepared by dissolving 25 mg of $K_2SO_4$ in 2 litre of water at 25$^\circ$C, assuming that it is completely dissociated.

Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K₂SO₄ in 2 litre of water at 25^°

Exercises · 1.41

K_2SO_4

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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 depends on the total particle concentration after dissociation. $K_2SO_4$ splits into three ions, tripling the effective molar concentration. The osmotic pressure is $\boxed{5.27 \times 10^{-3} \text{ atm}}$ .

Why osmotic pressure depends on particle count

Osmotic pressure measures the "push" exerted by solute particles trying to equalize concentration across a semipermeable membrane. The van 't Hoff equation tells us that osmotic pressure $\pi$ behaves like an ideal gas:

\pi = CRT

where $C$ is the molar concentration of particles, $R$ is the gas constant, and $T$ is absolute temperature.

The crucial insight: when an ionic compound dissolves and dissociates, each formula unit breaks into multiple ions. Each ion contributes independently to the osmotic pressure. So we need to account for the van 't Hoff factor $i$ , the number of particles produced per formula unit:

\pi = iCRT

For $K_2SO_4$ , complete dissociation gives:

K_2SO_4 \longrightarrow 2K^+ + SO_4^{2-}

That's three particles from one formula unit, so $i = 3$ .

Step-by-step calculation

1. Convert mass to moles

The molar mass of $K_2SO_4$ is:

M = 2(39) + 32 + 4(16) = 78 + 32 + 64 = 174 \text{ g/mol}

Given mass is $25 \text{ mg} = 0.025 \text{ g}$ , so:

n = \frac{0.025}{174} = 1.437 \times 10^{-4} \text{ mol}

2. Find the molar concentration of the solute

Volume is $2 \text{ L}$ , so:

C = \frac{1.437 \times 10^{-4}}{2} = 7.18 \times 10^{-5} \text{ mol/L}

3. Account for dissociation

Since $K_2SO_4$ produces $i = 3$ particles per formula unit, the effective particle concentration is:

$C_{\text{particles}} = i \times C = 3 \times 7.18 \times 10^{-5} = 2.154 \times 10^{-4} \text{ mol/L}$ …

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Common Mistakes & How to Avoid Them

Mistake 1: Forgetting the van't Hoff Factor ( $i$ )

The error: Students calculate osmotic pressure using $\pi = CRT$ directly, ignoring dissociation. For $K_2SO_4$ , which dissociates completely:

K_2SO_4 \rightarrow 2K^+ + SO_4^{2-}

This gives 3 ions per formula unit, so $i = 3$ .

How to avoid: Always check if the solute is ionic and whether it dissociates. For complete dissociation:

$i$ = number of ions produced per formula unit
For $K_2SO_4$ , $i = 3$ (not 1)

Use the correct formula: $\pi = iCRT$

Mistake 2: Unit Confusion (mg vs g, mL vs L)

The error: Using 25 mg as 25 g, or forgetting to convert volume to litres.

How to avoid: Always convert to standard SI units before plugging into formulas:

Mass: 25 mg = $25 \times 10^{-3}$ g
Volume: 2 L (already correct)
Molar mass of $K_2SO_4$ : $2(39.1) + 32 + 4(16) = 174.2$ g/mol

Quick check: Write units alongside every number in your calculation.

Mistake 3: Using Wrong Temperature Scale

The error: Plugging in $25^\circ C$ directly as $T$ without converting to Kelvin.

How to avoid: Always convert Celsius to Kelvin:

T(K) = 25 + 273 = 298 \text{ K}

Memory aid: "Gas constant R uses Kelvin — so must T."

Mistake 4: Incorrect Molarity Calculation

The error: Calculating moles correctly but then dividing by wrong volume or forgetting to convert mass to moles first.

Correct approach:

\text{Moles of } K_2SO_4 = \frac{25 \times 10^{-3} \text{ g}}{174.2 \text{ g/mol}} = 1.435 \times 10^{-4} \text{ mol}

\text{Molarity } C = \frac{1.435 \times 10^{-4} \text{ mol}}{2 \text{ L}} = 7.175 \times 10^{-5} \text{ M}

How to avoid: Write the formula step-by-step:

C = \frac{\text{mass (g)}}{\text{molar mass (g/mol)} \times \text{volume (L)}}

Mistake 5: Forgetting the Value of R …

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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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←Previous question1.40 — Determine the amount of CaCl₂ (i = 2.47) dissolved in 2.5 litre of wat…→Next section · answersAnswers to Intext Questions

Q.Determine the osmotic pressure of a solution prepared by dissolving 25 mg of $K_2SO_4$ in 2 litre of water at 25 $^\circ$ C, assuming that it is completely dissociated.

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 with Dissociation

Why osmotic pressure depends on particle count

Step-by-step calculation

Method: Van't Hoff Equation for Electrolyte Solutions

Step 1: Write the Van't Hoff equation

Step 2: Determine the Van't Hoff factor ( $i$ )

Step 3: Calculate moles of $K_2SO_4$

Step 4: Calculate molar concentration ( $C$ )

Step 5: Convert temperature to Kelvin …

Common Mistakes & How to Avoid Them

Mistake 1: Forgetting the van't Hoff Factor ( $i$ )

Mistake 2: Unit Confusion (mg vs g, mL vs L)

Mistake 3: Using Wrong Temperature Scale

Mistake 4: Incorrect Molarity Calculation

Mistake 5: Forgetting the Value of R …

Concept

Solution

Concept

Solution …

Concept

Solution …

Q.Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4K_2SO_4K2​SO4​ in 2 litre of water at 25∘^\circ∘C, assuming that it is completely dissociated.

Why osmotic pressure depends on particle count

Step-by-step calculation

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Q.Determine the osmotic pressure of a solution prepared by dissolving 25 mg of $K_2SO_4$ in 2 litre of water at 25 $^\circ$ C, assuming that it is completely dissociated.

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