The osmotic pressure exerted by a solution prepared by dissolving $$2.0$$ g of protein of molar mass $$60$$ kg mol$$^{-1}$$ in $$200$$ mL of water at $$27°$$C is ______ Pa. [integer value] (use $$R = 0.083$$ L bar mol$$^{-1}$$ K$$^{-1}$$)

We need to calculate the osmotic pressure of a protein solution.

The mass of the protein is 2.0 g, the molar mass is $$60\,\mathrm{kg\,mol^{-1}}$$ (equivalent to $$60000\,\mathrm{g\,mol^{-1}}$$), the volume of the solution is 200 mL (0.2 L), the temperature is 27 °C (300 K), and $$R = 0.083\,\mathrm{L\,bar\,mol^{-1}\,K^{-1}}$$.

The number of moles of protein is given by

$$n = \frac{2.0}{60000} = \frac{1}{30000}\text{ mol}.$$

The osmotic pressure is determined using

$$\pi = \frac{nRT}{V}$$

Substituting the values yields

$$\pi = \frac{\frac{1}{30000} \times 0.083 \times 300}{0.2}$$

This simplifies to

$$\pi = \frac{\frac{24.9}{30000}}{0.2} = \frac{0.00083}{0.2} = 0.00415\text{ bar}.$$

Since $$1\text{ bar} = 10^5\text{ Pa}$$, converting the pressure to pascals gives

$$\pi = 0.00415 \times 10^5 = 415\text{ Pa}.$$

Therefore, the osmotic pressure is 415 Pa.