Question

What is the molar mass of a solute when a solution is prepared by dissolving \(0.8\) g of solute in \(0.3\) dm\({}^{3}\) of water having osmotic pressure \(0.2\) atm at \(300\) K (R = \(0.082\) atm dm\({}^{3}\) K\({}^{-1}\) mol\({}^{-1}\))

• A. \(300\) g/mol
• B. \(280\) g/mol
• C. \(328\) g/mol
• D. \(345\) g/mol

Hint:

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

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Osmotic pressure follows \(\pi V = nRT\), where \(n = \dfrac{w}{M}\).

Step 2: Rearrange
\(M = \dfrac{wRT}{\pi V}\)

Step 3: Calculation
\(M = \dfrac{0.8 \times 0.082 \times 300}{0.2 \times 0.3} = \dfrac{19.68}{0.06} = 328\) g/mol

Step 4: Compare the options
The molar mass is \(328\) g/mol, option (C).

Final Answer:
The molar mass of the solute is 328 g/mol. \[ \boxed{\text{(C) }\text{328 g/mol}} \]