JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties
The osmotic pressure exerted by a solution prepared by dissolving \(2.0\, g\) of protein of molar mass \(60\, kg\, mol ^{-1}\) in \(200\, mL\) of water at \(27^{\circ} C\) is \(.....Pa\). [integer value] (use \(R =0.083 L\,bar\,mol ^{-1}\, K ^{-1}\) )
- A \(236\)
- B \(654\)
- C \(313\)
- D \(415\)
Answer & Solution
Correct Answer
(D) \(415\)
Step-by-step Solution
Detailed explanation
\(\pi= iCRT\) \(=\frac{1 \times 2}{60000 \times 0.2} \times 0.083 \times 300\) \(=0.00415\) bar \(\quad\left(\because 1\right.\) bar \(\left.=10^{5} Pa \right)\) So, \(0.00415 \times 10^{5} Pa =415 \,Pa\)
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