Q.An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?
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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 (Π).
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:
Molar mass=number of molesmass of solute (g)
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:
ΠV=nRT
where:
- Π = osmotic pressure (in atm or Pa)
- V = volume of solution (in L or m³)
- n = number of moles of solute
- R = ideal gas constant (0.0821 L·atm·mol⁻¹·K⁻¹ or 8.314 J·mol⁻¹·K⁻¹)
- T = absolute temperature (in K)
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:
Π=VnRT=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 Π at temperature T, you can find the molar mass M as follows:
- From ΠV=nRT, we get n=RTΠV
- But n=Mw (mass divided by molar mass)
- Equating: Mw=RTΠV
- Rearranging:
M=ΠVwRT
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.
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 (Π) 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:
Π=iCRT
Where:
- Π = 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:
Π=CRT
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:
PV=nRT⇒P=VnRT=CRT
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 Π=CRT 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:
Π=iCRT
Step 3: Linking to Molar Mass
We usually know mass of solute (w) and volume of solution (V). Molar concentration is:
C=Vn=Vw/M
where M = molar mass (g/mol).
Substitute into the osmotic pressure equation:
Π=i⋅MVw⋅RT
Rearrange to solve for molar mass:
M=ΠViwRT
This is the key formula used in experiments to find molar mass from osmotic pressure.
4. Why is this method special? …
Concept: Osmotic pressure is used here, but more directly this is a vapour-pressure lowering problem — the solute reduces the vapour pressure of water at its boiling point, and the observed pressure (1.004 bar) is the vapour pressure of the solution. At the normal boiling point of water (100°C), pure water exerts 1.013 bar.
Step 1 — Find the lowering of vapour pressure.
ΔP=P∘−P=1.013−1.004=0.009 bar
Step 2 — Apply Raoult’s law for a dilute solution.
For a non-volatile solute,
P∘ΔP=n1+n2n2≈n1n2 (since n2≪n1).
Here n2 = moles of solute, n1 = moles of solvent (water).
Step 3 — Express in terms of masses.
Mass of solute = 2 g per 100 g solution → solvent mass = 98 g. …
At the normal boiling point, the vapour pressure of pure water is 1.013 bar. The observed pressure drop (1.013 – 1.004 = 0.009 bar) is due to the solute. Using Raoult’s law for a dilute solution, the mole fraction of solute equals the relative lowering of vapour pressure, which gives the molar mass of the solute as approximately 41.3 g/mol.
Why this works
The normal boiling point of a liquid is the temperature at which its vapour pressure equals the external atmospheric pressure (1.013 bar for water at 100 °C). When a non-volatile solute is dissolved in water, the vapour pressure of the solution is lower than that of pure water at the same temperature — this is Raoult’s law in action.
The relative lowering of vapour pressure depends only on the mole fraction of the solute, not on its chemical identity. For a dilute solution, the mole fraction of solute is approximately the ratio of moles of solute to moles of solvent. Since we know the mass percentage of the solution, we can work backwards from the pressure drop to find the molar mass of the solute.
P0P0−P=xsolute=nsolute+nsolventnsolute
For dilute solutions, nsolute≪nsolvent, so:
P0P0−P≈nsolventnsolute
Step-by-step solution
1. Identify the given data
- Solvent: water (normal boiling point = 100 °C)
- Vapour pressure of pure water at 100 °C: P0=1.013 bar
- Vapour pressure of solution: P=1.004 bar
- Solution is 2% by mass of non-volatile solute → 2 g solute in 100 g solution, so mass of solvent (water) = 98 g.
2. Calculate the relative lowering of vapour pressure
P0P0−P=1.0131.013−1.004=1.0130.009≈0.008884
This dimensionless number equals the mole fraction of solute in the solution.
3. Express mole fraction in terms of moles
Let M be the molar mass of the solute (in g/mol).
Moles of solute: nsolute=M2
Moles of solvent (water, molar mass 18 g/mol): nsolvent=1898≈5.444 mol
Since the solution is dilute, nsolute≪nsolvent, so:
xsolute≈nsolventnsolute=5.4442/M
4. Equate and solve for M
5.4442/M=0.008884 …
Method: Relative Lowering of Vapour Pressure (Raoult's Law) for Molar Mass Determination
Key Concept
At the normal boiling point of a liquid, its vapour pressure equals the external atmospheric pressure. For water, this means the vapour pressure of pure water at 100 degrees C is 1.013 bar. When a non-volatile solute is dissolved, the vapour pressure of the solution at the same temperature (1.004 bar here) is lower -- this is Raoult's law, not osmotic pressure.
P∘P∘−P=xsolute=nsolute+nsolventnsolute≈nsolventnsolute(dilute solution)
Step-by-Step Solution
Step 1: Identify given data
- Solvent: water, normal boiling point = 100 degrees C, so P∘=1.013 bar
- Vapour pressure of the solution: P=1.004 bar
- 2% solute by mass -> 2 g solute per 100 g solution -> solvent (water) mass = 98 g
Step 2: Relative lowering of vapour pressure
P∘P∘−P=1.0131.013−1.004=1.0130.009≈0.00888
Step 3: Moles of solvent
nwater=1898≈5.444 mol
Step 4: Solve for the molar mass M of the solute
Let M be the molar mass of the solute; nsolute=M2. …
Here are the common mistakes students make on this exact type of problem, along with how to avoid each.
Mistake 1: Treating the given pressure as osmotic pressure
The error: The number "1.004 bar" looks like it could plug into π=CRT, so students reach for the osmotic-pressure formula.
Why it's wrong: The problem says the solution "exerts a pressure ... at the normal boiling point of the solvent." That is the vapour pressure of the solution at 100 degrees C, not an osmotic pressure -- there is no semipermeable membrane anywhere in this problem.
How to avoid:
- At the normal boiling point of a liquid, its vapour pressure equals the external (atmospheric) pressure -- for water that is 1.013 bar.
- The solution's vapour pressure at that same temperature is 1.004 bar, lower than pure water's because of the dissolved solute. This is Raoult's law / relative lowering of vapour pressure, not osmotic pressure.
Mistake 2: Forgetting to find the vapour pressure of pure water first
The error: Students try to use 1.004 bar directly without ever bringing in the vapour pressure of pure water at 100 degrees C (1.013 bar).
How to avoid:
- Always start with: Pwater at 100∘C∘=1.013 bar (this is just the external/atmospheric pressure, by the very definition of the normal boiling point).
- The lowering is: ΔP=P∘−P=1.013−1.004=0.009 bar.
Mistake 3: Using the mass of solution instead of the mass of solvent
The error: "2% solute" is read as "2 g solute + 100 g water," using 100 g as the solvent mass.
How to avoid:
- 2% by mass means 2 g of solute per 100 g of solution -- so the solvent (water) mass is 100−2=98 g, not 100 g.
Mistake 4: Forgetting the dilute-solution approximation
The error: Not knowing how to relate the relative lowering of vapour pressure to mole fraction.
How to avoid: …
- CBSE 2024Set 56/3/11 markMCQQ.A 1% solution of solute 'X' is isotonic with a 6% solution of sucrose (molar mass = 342 g mol−1). The molar mass of solute 'X' is : (A) 34·2 g mol−1 (B) 57 g mol−1 (C) 114 g mol−1 (D) 3·42 g 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 Π=iCRT, and since neither solute dissociates, i=1. So Π 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=molar mass (g mol⁻¹)mass of solute per litre (g L⁻¹).
For sucrose: Msucrose=34260 mol L⁻¹.
For X: MX=MX10 mol L⁻¹, where MX is the unknown molar mass in g mol⁻¹.
-
Set the molarities equal because the solutions are isotonic.
MX10=34260
- Solve for MX. Cross‑multiply: 10×342=60×MX
3420=60MX
MX=603420=57 …
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- CBSE 2023Set 56/1/11 markMCQQ.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 solutionSolution
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=100,000g mol−1. The molality is roughly 0.0001mol kg−1. Now compare the four colligative properties:
Property Proportionality constant Effect for m≈0.0001 Depression in freezing point Kf≈1.86K kg mol−1 (water) ΔTf≈0.0002K Elevation in boiling point Kb≈0.52K kg mol−1 (water) ΔTb≈0.00005K Relative lowering of vapour pressure p0Δp=χsolute ≈0.000002 Osmotic pressure π=CRT π≈2.5kPa at 298K 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 1–2K kg mol−1. For macromolecules, m∼10−4 yields ΔT∼10−4K, which requires extraordinarily sensitive thermometry.
-
Vapour-pressure lowering is proportional to mole fraction. For dilute solutions of high-molar-mass solutes, χsolute∼10−6, making the relative change in vapour pressure unmeasurable with ordinary equipment.
-
Osmotic pressure obeys π=CRT, where C is molar concentration. Even though C is small, the gas constant R=8.314J mol−1K−1 and room temperature T≈300K combine to give RT≈2500J mol−1=2500Pa L mol−1. A concentration of C=0.001mol L−1 produces π≈2.5kPa, easily measured with a simple manometer (a column height of about 25cm of water).
TipOsmotic 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 markMCQQ.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 Π=iMRT (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 Π 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 Π=iMRT, 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 outA 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: …
-
- CBSE 2023Set 56/3/11 markMCQQ.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.
The van’t Hoff equation for osmotic pressure:
Π=iMRT
where Π is osmotic pressure, i is the van’t Hoff factor (1 for non-electrolytes like most proteins), M is molarity (mol/L), R is the gas constant, and T is absolute temperature.
Step-by-step reasoning
-
Recall the four colligative properties
Relative lowering of vapour pressure (ΔP/P0), elevation in boiling point (ΔTb), depression in freezing point (ΔTf), and osmotic pressure (Π). 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=2×10−5 mol. The molarity is 2×10−4 M.
For boiling point elevation: ΔTb=Kb⋅m≈0.512×2×10−4≈1×10−4∘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 — also tiny.
For osmotic pressure: Π=MRT=(2×10−4)×0.0821×298≈0.0049 atm≈3.7 mm Hg. This is easily measurable with a simple column of mercury or water.
-
Compare the sensitivities
The key insight: ΔTb and ΔTf are proportional to molality, while Π is proportional to molarity. But the real difference is the magnitude of the constants. Kb and Kf 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 Kf and 100 times larger than Kb. This makes osmotic pressure the most sensitive colligative property by far.
-
Eliminate the other options …
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- CBSE 2020Set 56/3/11 markMCQQ.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
π=iCRT
where π 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 π∝C (at constant T and i).
Does (R) explain (A)?
Here's the subtle point. Saying "π 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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