JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties
At \(27^{\circ}\,C\), a solution containing \(2.5\,g\) of solute in \(250.0\,mL\) of solution exerts an osmotic pressure of \(400\,Pa\). The molar mass of the solute is \(..............g\) \(mol ^{-1}\) (Nearest integer)(Given : \(R =0.083\,L\,bar\,K ^{-1}\,mol ^{-1}\) )
- A \(62240\)
- B \(62258\)
- C \(62240\)
- D \(62250\)
Answer & Solution
Correct Answer
(D) \(62250\)
Step-by-step Solution
Detailed explanation
\(\pi= CRT\) \(\frac{400\,Pa }{10^5}=\frac{\frac{2.5\,g }{ M _o}}{250 / 1000\,L } \times 0.83 \frac{ L - bar }{ K \cdot mol } \times 300\,K\) \(M _0=62250\)
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