NEET · Chemistry · STD 12 -1. Solution and colligative properties
The following solutions were prepared by dissolving \(10\, \mathrm{~g}\) of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) in \(250 \,\mathrm{ml}\) of water \(\left(\mathrm{P}_{1}\right)\), \(10\, \mathrm{~g}\) of urea \(\left(\mathrm{CH}_{4} \mathrm{~N}_{2} \mathrm{O}\right)\) in \(250\, \mathrm{ml}\) of water \(\left(\mathrm{P}_{2}\right)\) and \(10\, \mathrm{~g}\) of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) in \(250\, \mathrm{ml}\) of water \(\left(\mathrm{P}_{3}\right)\). The right option for the decreasing order of osmotic pressure of these solutions is :
- A \(\mathrm{P}_{2}>\mathrm{P}_{1}>\mathrm{P}_{3}\)
- B \(\mathrm{P}_{1}>\mathrm{P}_{2}>\mathrm{P}_{3}\)
- C \(\mathrm{P}_{2}>\mathrm{P}_{3}>\mathrm{P}_{1}\)
- D \(\mathrm{P}_{3}>\mathrm{P}_{1}>\mathrm{P}_{2}\)
Answer & Solution
Correct Answer
(A) \(\mathrm{P}_{2}>\mathrm{P}_{1}>\mathrm{P}_{3}\)
Step-by-step Solution
Detailed explanation
\(\Pi=\mathrm{CRT}\) \(\Pi \propto \mathrm{C} \Rightarrow \Pi \propto \frac{1}{\mathrm{M}_{\mathrm{w}}}\)
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