2.5 g of protein containing only glycine (C$$_2$$H$$_5$$NO$$_2$$) is dissolved in water to make 500 mL of solution. The osmotic pressure of this solution at 300 K is found to be $$5.03 \times 10^{-3}$$ bar. The total number of glycine units present in the protein is
(Given: R = 0.083 L bar K$$^{-1}$$ mol$$^{-1}$$)

We are given that 2.5 g of a protein (composed entirely of glycine units) is dissolved in water to make 500 mL of solution, and the osmotic pressure at 300 K is $$5.03 \times 10^{-3}$$ bar.

Using the osmotic pressure formula $$\pi = cRT$$, we find $$c = \frac{\pi}{RT} = \frac{5.03 \times 10^{-3}}{0.083 \times 300} = \frac{5.03 \times 10^{-3}}{24.9} = 2.02 \times 10^{-4} \text{ mol/L}$$.

The number of moles of protein in 0.5 L is $$n = c \times V = 2.02 \times 10^{-4} \times 0.5 = 1.01 \times 10^{-4} \text{ mol}$$, and its molar mass is $$M = \frac{\text{mass}}{n} = \frac{2.5}{1.01 \times 10^{-4}} = 24752 \text{ g/mol}$$.

The molar mass of glycine ($$\text{C}_2\text{H}_5\text{NO}_2$$) is $$M_{\text{glycine}} = 2(12) + 5(1) + 14 + 2(16) = 24 + 5 + 14 + 32 = 75 \text{ g/mol}$$, and $$\text{Number of glycine units} = \frac{24752}{75} \approx 330$$.

Hence, the total number of glycine units present in the protein is 330.