© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the Relevant Property (Osmotic Pressure)
Osmotic pressure $ \pi $ is a colligative property given by the formula
$ \pi = i \, C \, R \, T, $
where $ C $ is the molar concentration (moles of solute per litre of solution), $ R $ is the gas constant, $ T $ is the temperature (in Kelvin), and $ i $ is the van't Hoff factor (which is 1 for non-electrolytes such as glucose, urea, and sucrose).
For solutions with the same volume of solvent, the molar concentration depends solely on the number of moles of solute. Therefore, lower the molar mass of the solute at a fixed mass, higher will be the number of moles, resulting in a higher osmotic pressure.
Step 2: Calculate or Compare Molar Masses of the Solutes
Glucose (C6H12O6): Molar mass ≈ 180 g/mol
Urea (CH4N2O): Molar mass ≈ 60 g/mol
Sucrose (C12H22O11): Molar mass ≈ 342 g/mol
Given the same mass (10 g) of each solute is dissolved, the number of moles is inversely proportional to its molar mass. Thus, for 10 g each:
Urea has the highest number of moles (because it has the smallest molar mass).
Glucose has the next highest number of moles.
Sucrose has the fewest moles (because it has the largest molar mass).
Step 3: Relate Molar Concentration to Osmotic Pressure
Since all solutions are made up to the same total volume (250 mL), the solution with the greatest number of moles of solute will have the highest molar concentration $ C $ and therefore the highest osmotic pressure $ \pi $. The order of osmotic pressures follows the order of moles:
$ \pi_{(\text{Urea})} > \pi_{(\text{Glucose})} > \pi_{(\text{Sucrose})}. $
Step 4: Match to the Given Labels (P1, P2, and P3)
• P1 is the solution of Glucose
• P2 is the solution of Urea
• P3 is the solution of Sucrose
The decreasing order of their osmotic pressure is:
$ P_{2} ($Urea$) > P_{1} ($Glucose$) > P_{3} ($Sucrose$). $
Final Answer
The correct decreasing order of osmotic pressure is:
P2 > P1 > P3
