At 27°C, a solution containing 2.5 g of solute in 250.0 mL of solution exerts an osmotic pressure of 400 Pa. The molar mass of the solute is g mol$$^{-1}$$ (Nearest integer)
(Given: R = 0.083 L bar$$^{-1}$$ mol$$^{-1}$$)

Using the osmotic pressure formula:

$$ \pi = \frac{n}{V}RT = \frac{w}{MV}RT $$

Rearranging for molar mass $$M$$:

$$ M = \frac{wRT}{\pi V} $$

Given: $$w = 2.5$$ g, $$V = 250.0$$ mL $$= 0.25$$ L, $$T = 27°C = 300$$ K, $$R = 0.083$$ L bar K$$^{-1}$$ mol$$^{-1}$$

Converting osmotic pressure to bar: $$\pi = 400$$ Pa $$= \frac{400}{10^5}$$ bar $$= 0.004$$ bar

$$ M = \frac{2.5 \times 0.083 \times 300}{0.004 \times 0.25} = \frac{62.25}{0.001} = 62250 \text{ g mol}^{-1} $$