Question
Solution A is prepared by dissolving 1 g of a protein (molar mass $= 50000\text{ }g{\text{ }mol}^{- 1}$ ) 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 :
( $R = 0.083\text{ }L$ bar ${mol}^{- 1}{\text{ }K}^{- 1}$ )
Options
Solution
Sol. [Key Concept: $\pi = CRT$ for each solution; on mixing use total moles over total volume.]
$\ \ ^{'}x^{'} = \frac{1}{50,000 \times 0.5} \times R \times 300 = 9.96 \times 10^{- 4}$ bar
$\ ^{'}\mathbf{y}^{'} = \frac{2}{50,000 \times 1} \times R \times 300 = 9.96 \times 10^{- 4}$ bar
$\ ^{'}\mathbf{z}^{'} = \frac{\ ^{'}x^{'} \times \frac{1}{2} + \ ^{'}y^{'} \times 1}{\frac{3}{2}} = 9.96 \times 10^{- 4}$ bar
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