Q.200 cm$^3$ 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 $2.57 \times 10^{-3}$ bar. Calculate the molar mass of the protein.

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 ($\Pi$) of a dilute solution is given by the van't Hoff equation:

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.

1. Rearrange the formula to solve for the molar mass ($M$): $M = \frac{w R T}{\Pi V}$
2. Convert the given volume from cm$^3$ to Litres: $V = 200 \text{ cm}^3 = 0.200 \text{ L}$
3. Substitute the given values. Use $R = 0.083 \text{ L bar K}^{-1} \text{ mol}^{-1}$ — the value NCERT's own Solution uses, consistent with the pressure unit (bar): …

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:

1. 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.
2. 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:

Let's break down the calculation step-by-step.

1. Identify the given information and the target:

   • Volume of solution ($V$) = $200 \text{ cm}^3$
   • Mass of protein ($w$) = $1.26 \text{ g}$
   • Temperature ($T$) = $300 \text{ K}$
   • Osmotic pressure ($\Pi$) = $2.57 \times 10^{-3} \text{ bar}$
   • We need to calculate the molar mass of the protein ($M$).

2. Ensure consistent units:

   Since the pressure is given in bar and we will convert volume to litres, we use the gas constant $R = 0.083 \text{ L bar mol}^{-1} \text{ K}^{-1}$ — the value NCERT's own Solution uses throughout this chapter.

   • Convert volume from $\text{cm}^3$ to Litres: $V = 200 \text{ cm}^3 = 200 \times 10^{-3} \text{ L} = 0.200 \text{ L}$
   • All other units are already consistent with the chosen $R$ value.

3. 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 ($V$):

   $C = \frac{n}{V}$

   The number of moles ($n$) can also be expressed as the mass of the solute ($w$) divided by its molar mass ($M$):

   $n = \frac{w}{M}$

   Substituting this into the expression for $C$:

   $C = \frac{w/M}{V} = \frac{w}{MV}$

4. Substitute the expression for $C$ into the van't Hoff equation:

   $\Pi = \left(\frac{w}{MV}\right)RT$

5. Rearrange the equation to solve for molar mass ($M$):

   $M = \frac{wRT}{\Pi V}$ …

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

For a non-electrolyte (like a protein), $i = 1$, so:

Where:

• $\pi$ = 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

Step 2: Rearrange the formula to solve for $M$

Step 3: Substitute the given values

• $w = 1.26 \, \text{g}$
• $R = 0.083 \, \text{L·bar·mol}^{-1}\text{·K}^{-1}$
• $T = 300 \, \text{K}$
• $\pi = 2.57 \times 10^{-3} \, \text{bar}$
• $V = 0.200 \, \text{L}$

Step 4: Calculate numerator

Step 5: Calculate denominator

Step 6: Final calculation

$M = \frac{31.374}{5.14 \times 10^{-4}} \approx 6.10 \times 10^4 \, \text{g/mol}$ …

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 $\pi = iCRT$ uses concentration $C$ in mol/L. Volume must be in litres, not cm³.

• Wrong: $200 \, \text{cm}^3$ used as $200 \, \text{L}$ or $200 \, \text{mL}$ without conversion.
• Correct: $200 \, \text{cm}^3 = 0.200 \, \text{L}$.

How to avoid: Always write the conversion step:

$V = 200 \, \text{cm}^3 \times \frac{1 \, \text{L}}{1000 \, \text{cm}^3} = 0.200 \, \text{L}$.

Mistake 2: Using the wrong value of $R$

The gas constant $R$ must match the units of pressure and volume.

• Given: $\pi = 2.57 \times 10^{-3} \, \text{bar}$, $T = 300 \, \text{K}$.
• Wrong: Using $R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1}$ without converting bar to atm.
• Correct: Use a bar-based $R$. NCERT's Solution uses $R = 0.083 \, \text{L bar mol}^{-1} \text{K}^{-1}$ — using the fuller $0.08314$ instead shifts the third significant figure of the answer (≈ $6.11 \times 10^4$ vs ≈ $6.10 \times 10^4$), 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 $R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1}$ 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 $i = \text{(number of ions)}$.
• 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 $\pi = \frac{w}{M V} RT$ (with $i=1$) is:

• Wrong: Writing $M = \frac{w \pi}{RT V}$ or inverting the fraction.
• Correct: $M = \frac{1.26 \, \text{g} \times 0.083 \, \text{L bar mol}^{-1} \text{K}^{-1} \times 300 \, \text{K}}{2.57 \times 10^{-3} \, \text{bar} \times 0.200 \, \text{L}}$

How to avoid:

Write the formula step by step:

$\pi = \frac{n}{V} RT = \frac{w}{M V} RT$

Then solve for $M$: …