Question

Calculate the mass of nonvolatile solute dissolved in \(0.3\) dm\(^3\) water having osmotic pressure \(0.1\) atm at \(300\)K.
[Molar mass of solute = \(328\) g mol\(^{-1}\), R = \(0.082\) dm\(^3\)atm K\(^{-1}\)mol\(^{-1}\)]

• A. \(0.4\) g
• B. \(0.6\) g
• C. \(0.8\) g
• D. \(1.0\) g

Hint:

Use pi = CRT with C in mol per dm3, then multiply moles by molar mass.

Answer 1

The Correct Option is A

Step 1: Get the concentration:
$C = \pi/(RT) = 0.1/(0.082 \times 300) = 0.1/24.6 = 4.065 \times 10^{-3}$ mol dm$^{-3}$.

Step 2: Get the moles:
Volume is 0.3 dm$^3$, so $n = 4.065 \times 10^{-3} \times 0.3 = 1.22 \times 10^{-3}$ mol.

Step 3: Convert to mass:
$1.22 \times 10^{-3} \times 328 = 0.40$ g.

Final Answer:
The mass dissolved is 0.4 g, option (A). \[ \boxed{0.4 \text{ g}} \]