Solution A is prepared by dissolving 1 g of a protein (molar mass = 50000 g mol$$^{-1}$$) in 0.5 L of water at 300 K. Its osmotic pressure is $$x$$ bar. Solution B is made by dissolving 2 g of same protein in 1 L of water at 300 K. Osmotic pressure of solution B is $$y$$ bar. Entire solution of A is mixed with entire solution of B at same temperature. The osmotic pressure of resultant solution is $$z$$ bar. $$x$$, $$y$$ and $$z$$ respectively are :
($$R = 0.083$$ L bar mol$$^{-1}$$ K$$^{-1}$$)

The osmotic pressure for dilute non-electrolyte solutions is given by van ’t Hoff’s equation: $$\pi = C\,R\,T$$, where $$\pi$$ is the osmotic pressure (bar), $$C$$ is the molarity (mol L$$^{-1}$$), $$R = 0.083$$ L bar mol$$^{-1}$$ K$$^{-1}$$ and $$T = 300$$ K.

Solution A
Mass of protein dissolved $$= 1$$ g, molar mass $$= 50000$$ g mol$$^{-1}$$. Number of moles $$n_A = \frac{1}{50000} = 2.0 \times 10^{-5}\text{ mol}$$. Volume $$V_A = 0.5$$ L. Molarity $$C_A = \frac{n_A}{V_A} = \frac{2.0 \times 10^{-5}}{0.5} = 4.0 \times 10^{-5}\text{ M}$$.

Osmotic pressure of A:
$$\pi_A = C_A R T = \left(4.0 \times 10^{-5}\right)\left(0.083\right)\left(300\right)$$ $$\pi_A = 9.96 \times 10^{-4}\text{ bar}$$. Hence $$x = 9.96 \times 10^{-4}\text{ bar}$$.

Solution B
Mass of protein dissolved $$= 2$$ g. Number of moles $$n_B = \frac{2}{50000} = 4.0 \times 10^{-5}\text{ mol}$$. Volume $$V_B = 1.0$$ L. Molarity $$C_B = \frac{n_B}{V_B} = \frac{4.0 \times 10^{-5}}{1.0} = 4.0 \times 10^{-5}\text{ M}$$.

Osmotic pressure of B:
$$\pi_B = C_B R T = \left(4.0 \times 10^{-5}\right)\left(0.083\right)\left(300\right)$$ $$\pi_B = 9.96 \times 10^{-4}\text{ bar}$$. Hence $$y = 9.96 \times 10^{-4}\text{ bar}$$.

Resultant solution after mixing A and B
Total moles $$n_{\text{total}} = n_A + n_B = 2.0 \times 10^{-5} + 4.0 \times 10^{-5} = 6.0 \times 10^{-5}\text{ mol}$$. Total volume $$V_{\text{total}} = V_A + V_B = 0.5 + 1.0 = 1.5\text{ L}$$.

Molarity of mixture:
$$C_{\text{mix}} = \frac{n_{\text{total}}}{V_{\text{total}}} = \frac{6.0 \times 10^{-5}}{1.5} = 4.0 \times 10^{-5}\text{ M}$$.

Osmotic pressure of the mixture:
$$\pi_{\text{mix}} = C_{\text{mix}} R T = \left(4.0 \times 10^{-5}\right)\left(0.083\right)\left(300\right)$$ $$\pi_{\text{mix}} = 9.96 \times 10^{-4}\text{ bar}$$. Therefore $$z = 9.96 \times 10^{-4}\text{ bar}$$.

Thus $$x = y = z = 9.96 \times 10^{-4}\text{ bar}$$, which corresponds to

Option A which is: $$9.96 \times 10^{-4}$$; $$9.96 \times 10^{-4}$$; $$9.96 \times 10^{-4}$$.