20 g of a substance were dissolved in 500 mL of water and the osmotic pressure of the solution was found to be 600 mm of mercury at 15°C. The molecular weight of a substance is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 1198
The osmotic pressure of a dilute solution gives the molar mass of the solute through ΠV = (w2/M2)RT:
\( \displaystyle M_2 = \frac{w_2 R T}{\Pi V} \)
\( \displaystyle w_2 = 20\ \mathrm{g},\quad V = 0.5\ \mathrm{L},\quad T = 15 + 273 = 288\ \mathrm{K},\quad R = 0.0821\ \mathrm{L\,atm\,K^{-1}\,mol^{-1}} \)
\( \displaystyle \Pi = \frac{600}{760}\ \mathrm{atm} = 0.789\ \mathrm{atm} \)
\( \displaystyle M_2 = \frac{20 \times 0.0821 \times 288}{0.789 \times 0.5} = \frac{472.9}{0.3947} = 1198\ \mathrm{g\,mol^{-1}} \)
So the molar mass of the substance is 1198 g mol−1.
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:
Π = C R T (1.39)
Here Π is the osmotic pressure and R is the gas constant.
Π = (n2/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,
ΠV = w2RT/M2 (1.41)
or M2 = w2RT/ΠV (1.42)
Thus, knowing the quantities w2, T, Π and V we can calculate the molar mass of the solute.”