Q.200 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 2.57×10−3 bar. Calculate the molar mass of the protein.
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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? …
The key idea is using Osmotic Pressure to determine Molar Mass. Osmotic pressure is a colligative property, particularly useful for finding the molar masses of macromolecules like proteins, as it yields large, easily measurable values even for dilute solutions.
The osmotic pressure (Π) of a dilute solution is given by the van't Hoff equation:
Π=VnRT=MVwRT
Here, w is the mass of the solute, M is its molar mass, V is the volume of the solution in Litres, R is the gas constant, and T is the temperature in Kelvin.
- Rearrange the formula to solve for the molar mass (M): M=ΠVwRT
- Convert the given volume from cm3 to Litres: V=200 cm3=0.200 L
- Substitute the given values. Use R=0.083 L bar K−1 mol−1 — the value NCERT's own Solution uses, consistent with the pressure unit (bar): …
Osmotic pressure is a colligative property used to determine the molar mass of macromolecules like proteins. Applying the van't Hoff equation with the book's R=0.083 L bar K−1mol−1 gives the molar mass of the protein as 61,022 g mol−1 (≈ 6.10×104 g mol−1).
Proteins are large molecules, often referred to as macromolecules. Determining their molar mass is crucial for understanding their structure and function. Traditional colligative properties like elevation in boiling point or depression in freezing point are often not suitable for macromolecules for a few key reasons:
- Small Magnitude: For a given mass concentration, the molar concentration (and thus the colligative effect) is very small due to the large molar mass. This makes the changes in boiling or freezing points difficult to measure accurately.
- Temperature Sensitivity: Proteins are often sensitive to temperature changes and can denature (lose their natural structure and function) at high or low temperatures, making boiling or freezing point measurements impractical.
Osmotic pressure, however, offers a distinct advantage. Even for very dilute solutions of macromolecules, the osmotic pressure can be significant and accurately measurable at physiological temperatures. This makes it the preferred method for determining the molar masses of polymers and proteins.
The relationship between osmotic pressure and molar concentration for dilute solutions is given by the van't Hoff equation, which is analogous to the ideal gas equation:
Π=CRT
Where:
Π is the osmotic pressure
C is the molar concentration (molarity) of the solute
R is the gas constant
T is the temperature in Kelvin
Let's break down the calculation step-by-step.
-
Identify the given information and the target:
- Volume of solution (V) = 200 cm3
- Mass of protein (w) = 1.26 g
- Temperature (T) = 300 K
- Osmotic pressure (Π) = 2.57×10−3 bar
- We need to calculate the molar mass of the protein (M).
-
Ensure consistent units:
Since the pressure is given in bar and we will convert volume to litres, we use the gas constant R=0.083 L bar mol−1 K−1 — the value NCERT's own Solution uses throughout this chapter.
- Convert volume from cm3 to Litres: V=200 cm3=200×10−3 L=0.200 L
- All other units are already consistent with the chosen R value.
-
Relate molar concentration to molar mass:
The molar concentration (C) is defined as the number of moles of solute (n) per unit volume of solution (V):
C=Vn
The number of moles (n) can also be expressed as the mass of the solute (w) divided by its molar mass (M):
n=Mw
Substituting this into the expression for C:
C=Vw/M=MVw
-
Substitute the expression for C into the van't Hoff equation:
Π=(MVw)RT
-
Rearrange the equation to solve for molar mass (M):
M=ΠVwRT …
Method: Osmotic Pressure Method for Molar Mass Determination
This method uses the van't Hoff equation for osmotic pressure, which relates the colligative property to the molar concentration of the solute.
Key Formula
π=i⋅C⋅R⋅T
For a non-electrolyte (like a protein), i=1, so:
π=Vn⋅R⋅T=M⋅Vw⋅R⋅T
Where:
- π = osmotic pressure (in bar)
- w = mass of solute (in g)
- M = molar mass (in g/mol)
- V = volume of solution (in L)
- R = gas constant (0.083 L·bar·mol−1·K−1 — the value NCERT uses in this chapter)
- T = temperature (in K)
Step-by-Step Solution
Step 1: Convert volume to litres
V=200cm3=0.200L
Step 2: Rearrange the formula to solve for M
M=π⋅Vw⋅R⋅T
Step 3: Substitute the given values
- w=1.26g
- R=0.083L⋅bar⋅mol−1⋅K−1
- T=300K
- π=2.57×10−3bar
- V=0.200L
M=2.57×10−3×0.2001.26×0.083×300
Step 4: Calculate numerator
1.26×0.083×300=31.374
Step 5: Calculate denominator
2.57×10−3×0.200=5.14×10−4
Step 6: Final calculation
M=5.14×10−431.374≈6.10×104g/mol …
Here are the common mistakes students make on this exact problem, along with how to avoid each one.
Mistake 1: Forgetting to convert volume to litres
The osmotic pressure formula π=iCRT uses concentration C in mol/L. Volume must be in litres, not cm³.
- Wrong: 200cm3 used as 200L or 200mL without conversion.
- Correct: 200cm3=0.200L.
How to avoid: Always write the conversion step:
V=200cm3×1000cm31L=0.200L.
Mistake 2: Using the wrong value of R
The gas constant R must match the units of pressure and volume.
- Given: π=2.57×10−3bar, T=300K.
- Wrong: Using R=0.0821L atm mol−1K−1 without converting bar to atm.
- Correct: Use a bar-based R. NCERT's Solution uses R=0.083L bar mol−1K−1 — using the fuller 0.08314 instead shifts the third significant figure of the answer (≈ 6.11×104 vs ≈ 6.10×104), so quote the book's value in exams.
How to avoid:
Check the pressure unit in the question. If it's in bar, use the bar-based R (0.083, as in the textbook).
If it's in atm, use R=0.0821L atm mol−1K−1 and convert pressure first.
Mistake 3: Forgetting the van't Hoff factor i for a non-electrolyte
Proteins in water are non-electrolytes — they do not dissociate.
- Wrong: Assuming i=2 or i=(number of ions).
- Correct: For a protein, i=1.
How to avoid:
Identify the solute type. If it's a non-electrolyte (like protein, sugar, urea), set i=1. Only use i>1 for salts that dissociate.
Mistake 4: Mixing up the formula for molar mass
The correct rearrangement of π=MVwRT (with i=1) is:
M=πVwRT
- Wrong: Writing M=RTVwπ or inverting the fraction.
- Correct: M=2.57×10−3bar×0.200L1.26g×0.083L bar mol−1K−1×300K
How to avoid:
Write the formula step by step:
π=VnRT=MVwRT
Then solve for M: …
- 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 …
-
- 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. …
-
- 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 …
-
- 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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