JEE MAIN - Chemistry (2024 - 29th January Morning Shift - No. 30)

The osmotic pressure of a dilute solution is $$7 \times 10^5 \mathrm{~Pa}$$ at $$273 \mathrm{~K}$$. Osmotic pressure of the same solution at $$283 \mathrm{~K}$$ is _________ $$\times 10^4 \mathrm{Nm}^{-2}$$.
Answer
73

Explanation

To calculate the osmotic pressure of the solution at the new temperature, we can use the formula:

$ \pi = \text{CRT} $

Since the concentration $ C $ and the gas constant $ R $ remain constant, the ratio of the osmotic pressures at two different temperatures is given by:

$ \frac{\pi_1}{\pi_2} = \frac{T_1}{T_2} $

Thus, we can express the osmotic pressure at the second temperature as:

$ \pi_2 = \frac{\pi_1 \cdot T_2}{T_1} $

Substituting the given values:

$ \pi_2 = \frac{7 \times 10^5 \times 283}{273} $

After calculation, this simplifies to:

$ \pi_2 = 72.56 \times 10^4 \, \text{Nm}^{-2} $

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