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: Understanding the Concept:
Osmotic pressure of a dilute solution of a nonelectrolyte follows the van 't Hoff equation, which looks like the ideal gas law.

Step 2: Key Formula or Approach:
\[ \pi V = nRT = \frac{w}{M}RT \;\Rightarrow\; M = \frac{wRT}{\pi V} \]

Step 3: Detailed Explanation:
Given \(w = 6\) g, \(V = 1 \text{ dm}^3\), \(\pi = 2.4\) atm, \(T = 300\) K, \(R = 0.0821\).
\[ M = \frac{6\times0.0821\times300}{2.4\times1} \]
\[ 6\times0.0821 = 0.4926, \qquad 0.4926\times300 = 147.78 \]
\[ M = \frac{147.78}{2.4} = 61.58 \text{ g mol}^{-1} \]
The other options do not result from these values. For example 74.12 would need a pressure near 2.0 atm.

Final Answer:
The molar mass of the solute is 61.58 g/mol, option (C). \[ \boxed{61.58 \text{ g mol}^{-1} \text{ (C)}} \]