Question

\(40\) gram nonelectrolyte solute having molar mass \(180 \text{g mol}^{-1}\) dissolved in water has osmotic pressure \(2 \text{atm}\) at \(300 \text{K}\). Calculate the volume of solution. (\(\text{R} = 0.0821 \text{atm mol}^{-1}\text{K}^{-1}\))

• A. \(2.10 \text{dm}^3\)
• B. \(2.34 \text{dm}^3\)
• C. \(2.74 \text{dm}^3\)
• D. \(3.40 \text{dm}^3\)

Hint:

Use pi V = n R T with n found from mass and molar mass.

Answer 1

The Correct Option is C

Step 1: Rearrange to concentration:
$\pi = CRT$, so $C = \frac{\pi}{RT} = \frac{2}{0.0821 \times 300} = 0.0812$ mol/L.

Step 2: Volume:
Moles of solute $= \frac{40}{180} = 0.2222$ mol.
$V = \frac{n}{C} = \frac{0.2222}{0.0812} = 2.74$ L $= 2.74$ dm$^3$.

Final Answer:
$V = 2.74$ dm$^3$, option (C). \[ \boxed{2.74\ \text{dm}^3} \]