Question

Calculate the molar mass of nonelectrolyte solute when 6 gram of it is dissolved in 1 \(\text{dm}^3\) water has osmotic pressure \(2.4 \text{atm}\) at 300 K \((R = 0.0821 \text{atm dm}^3 \text{K}^{-1} \text{mol}^{-1})\)

• A. \(45.0 \text{g mol}^{-1}\)
• B. \(74.12 \text{g mol}^{-1}\)
• C. \(61.58 \text{g mol}^{-1}\)
• D. \(90.28 \text{g mol}^{-1}\)

Hint:

Use pi = (w/M)(RT/V) and solve for M.

Answer 1

The Correct Option is C

Step 1: Find the moles from pressure:
$n = \pi V/(RT) = (2.4\times1)/(0.0821\times300) = 2.4/24.63 = 0.09745$ mol.

Step 2: Molar mass:
$M = \text{mass}/n = 6/0.09745 = 61.57$ g/mol.

Step 3: Compare:
This matches option (C), 61.58 g/mol, within rounding. The sample is 6 g for about 0.097 mol, so each mole weighs about 61.6 g.

Final Answer:
$M \approx 61.58$ g/mol, option (C). \[ \boxed{61.58 \text{ g mol}^{-1} \text{ (C)}} \]