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: Understanding the Concept:
For a dilute solution of a nonelectrolyte, the osmotic pressure follows the van t Hoff equation \(\pi V = nRT\), where \(n\) is the number of moles of solute.

Step 2: Key Formula or Approach:
\(n = \frac{w}{M}\) and \(V = \frac{nRT}{\pi}\).

Step 3: Detailed Explanation:
\(n = \frac{40}{180} = 0.2222\) mol.
\[ V = \frac{0.2222 \times 0.0821 \times 300}{2} = \frac{5.473}{2} = 2.74\ \text{L} \]
One litre equals one dm\(^3\), so \(V = 2.74\) dm\(^3\).
The other values come from using a wrong temperature or a wrong molar mass, and so they do not match.

Final Answer:
The volume is \(2.74\) dm\(^3\), option (C). \[ \boxed{2.74\ \text{dm}^3} \]