TS EAMCET · Chemistry · Solutions
The osmotic pressure (in atm) of an aqueous solution containing 0.01 mol of NaCl (degree of dissociation 0.94 ) and 0.03 mol of glucose in 500 mL at \(27^{\circ} \mathrm{C}\) is \(\left(\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)\)
- A \(2.43\)
- B \(4.23\)
- C \(3.24\)
- D \(3.42\)
Answer & Solution
Correct Answer
(A) \(2.43\)
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