JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties
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\(^{-1})\)
- A \(9.96 \times 10^{-4}; 9.96 \times 10^{-4}; 9.96 \times 10^{-4}\)
- B \(9.96 \times 10^{-4}; 9.96 \times 10^{-4}; 19.92 \times 10^{-4}\)
- C \(4.98 \times 10^{-4}; 4.98 \times 10^{-4}; 9.96 \times 10^{-4}\)
- D \(4.98 \times 10^{-4}; 4.98 \times 10^{-4}; 4.98 \times 10^{-4}\)
Answer & Solution
Correct Answer
(A) \(9.96 \times 10^{-4}; 9.96 \times 10^{-4}; 9.96 \times 10^{-4}\)
Step-by-step Solution
Detailed explanation
Osmotic pressure is given by \(\pi = CRT = \dfrac{w}{MV}RT\) For solution A: \(x = \dfrac{1}{50000 \times 0.5} \times 0.083 \times 300\) \(x = \dfrac{1}{25000} \times 24.9 = 9.96 \times 10^{-4}\) bar For solution B: \(y = \dfrac{2}{50000 \times 1} \times 0.083 \times 300\)…
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