Question

A solution is prepared by dissolving 0.3 g of a non-volatile non-electrolyte solute A of molar mass 60 g mol$^{-1}$ and 0.9 g of a non-volatile non-electrolyte solute B of molar mass 180 g mol$^{-1}$ in 100 mL H$_2$O at 27$^\circ$C. Osmotic pressure of the solution will be
[Given: R = 0.082 L atm K$^{-1}$ mol$^{-1}$]

• A. 1.23 atm
• B. 0.82 atm
• C. 2.46 atm
• D. 1.47 atm
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Hint:

For non-electrolytes, osmotic pressure depends only on the total number of solute particles present in the solution.

Answer 1

The Correct Option is C

Step 1: Calculate moles of solute A.
Moles of A $= \dfrac{0.3}{60} = 0.005$ mol
Step 2: Calculate moles of solute B.
Moles of B $= \dfrac{0.9}{180} = 0.005$ mol
Step 3: Calculate total moles of solute.
Total moles $= 0.005 + 0.005 = 0.01$ mol
Step 4: Apply osmotic pressure formula.
\[ \pi = \dfrac{nRT}{V} \] \[ \pi = \dfrac{0.01 \times 0.082 \times 300}{0.1} \] \[ \pi = 2.46 \text{ atm} \]
Step 5: Conclusion.
The osmotic pressure of the solution is 2.46 atm.