Target Exam

CUET

Subject

Chemistry

Chapter

Physical: Solutions

Question:

Which of the following colligative property is used to calculate the molar mass of biomolecule?

Options:

Relative lowering of vapour pressure

Elevation in boiling point

Depression in freezing point

Osmotic pressure

Correct Answer:

Osmotic pressure

Explanation:

The correct answer is option 4. Osmotic pressure.

Colligative properties are those properties of solutions that depend on the number of solute particles in a given amount of solvent, rather than the nature of the solute particles. The key colligative properties are:

I. Relative Lowering of Vapor Pressure

II. Elevation in Boiling Point

III. Depression in Freezing Point

IV. Osmotic Pressure

Osmotic Pressure*(\(\pi\)) is the pressure required to prevent the flow of solvent into a solution through a semipermeable membrane. It is given by the formula:

\(\pi = \frac{n}{V}RT \)

where:

\(\pi\) is the osmotic pressure,

\(n\) is the number of moles of solute,

\(V\) is the volume of the solution,

\(R\) is the gas constant (0.0821 L·atm·K⁻¹·mol⁻¹),

\(T\) is the temperature in Kelvin.


We need to measure the osmotic pressure (\(\pi\)) of the solution. This can be done using an osmometer, which measures the pressure exerted by the solute particles in the solution.

We have to then determine solution parameters:

Measure the volume of the solution (\(V\)).

Record the temperature (\(T\)) of the solution.

Rearranging the formula for osmotic pressure to solve for the number of moles of solute (\(n\)):

\(n = \frac{\pi V}{RT}\)

The molar mass \(M\) can then be calculated using:

\(M = \frac{m}{n} \)

where \(m\) is the mass of the solute.

Substituting \(n\) from the osmotic pressure formula:

\(M = \frac{m \cdot RT}{\pi V} \)

Example

Suppose we have a solution with an osmotic pressure of 2.0 atm at 298 K, and you know the volume of the solution is 1.0 L. If you have 10 grams of the biomolecule in the solution, you can calculate the molar mass as follows:

\( n = \frac{\pi V}{RT} \)

Substituting the values:

\(n = \frac{2.0 \text{ atm} \times 1.0 \text{ L}}{0.0821 \text{ L·atm·K⁻¹·mol⁻¹} \times 298 \text{ K}} \approx 0.082 \text{ mol} \)

\( M = \frac{m}{n} = \frac{10 \text{ g}}{0.082 \text{ mol}} \approx 122 \text{ g/mol} \)

Conclusion

Osmotic pressure is effective for calculating the molar mass of biomolecules because it provides a direct way to relate the amount of solute in the solution to measurable physical properties. This is especially useful for large biomolecules where other colligative properties might be less practical due to the small changes they produce.