The osmotic pressure of a dilute solution is $$7 \times 10^5$$ Pa at $$273$$ K. Osmotic pressure of the same solution at $$283$$ K is _______ $$\times 10^4$$ Nm$$^{-2}$$. (Nearest integer)

We need to find the osmotic pressure of a dilute solution at 283 K, given its osmotic pressure at 273 K.

Recall the relationship between osmotic pressure and temperature.

For a dilute solution, osmotic pressure is given by the van't Hoff equation:

$$\pi = CRT = \frac{n}{V}RT$$

where $$C$$ is the molar concentration, $$R$$ is the gas constant, and $$T$$ is the absolute temperature.

For the same solution (same $$C$$): $$\pi \propto T$$

Set up the proportion.

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

$$\pi_2 = \pi_1 \times \frac{T_2}{T_1} = 7 \times 10^5 \times \frac{283}{273}$$

Calculate.

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

$$= \frac{1981}{273} \times 10^5 = 7.2564 \times 10^5 \text{ Pa}$$

$$= 72.564 \times 10^4 \text{ Nm}^{-2} \approx 73 \times 10^4 \text{ Nm}^{-2}$$

The answer is 73.