The correct relation for determining the molar mass of solute ( M_2 )…

The correct answer is Option (2) → \(M_2=\frac{w_2RT}{\Pi V}\)

Putting \( \Pi = \frac{n_2}{V}\,RT \) and \( n_2 = w_2/M_2 \) together:

    \( \displaystyle M_2 = \frac{w_2 RT}{\Pi V} \)

So the correct relation is \( M_2 = \frac{w_2RT}{\Pi V} \). The method is widely used for proteins and polymers, because the measurement is made at room temperature and even a very dilute solution gives a measurable osmotic pressure.

Shortcut: Check the units. With w₂ in g, R in L bar K⁻¹ mol⁻¹, T in K, \( \Pi \) in bar and V in L, only Option (2) gives g mol⁻¹. Option (1) gives g² mol⁻¹, Option (3) has an extra K, and Option (4) has an extra L^1/2.

NCERT (2026-27 edition):
Osmosis and Osmotic Pressure – “… For dilute solutions, it has been found experimentally that osmotic pressure is proportional to the molarity, C of the solution at a given temperature T. Thus:

    \( \Pi = C\,R\,T \)   (1.39)

…

    \( \Pi = (n_2/V)\,R\,T \)   (1.40)

Here V is volume of a solution in litres containing n₂ moles of solute. If w₂ grams of solute, of molar mass, M₂ is present in the solution, then n₂ = w₂/M₂ and we can write,

    \( \displaystyle \Pi V = \frac{w_2\,R\,T}{M_2} \)   (1.41)

    or \( \displaystyle M_2 = \frac{w_2\,R\,T}{\Pi V} \)   (1.42)”