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