Question

The 'x' g nonvolatile solute having molar mass 196 gmol\(^{-1}\) is dissolved in 3 dm\(^3\) water. Calculate the value of 'x' if resulting solution has osmotic pressure 0.8 atm at 300 K (\(R = 0.0821\) dm\(^3\) atm K\(^{-1}\) mol\(^{-1}\))

• A. 19.1 g
• B. 9.8 g
• C. 14.7 g
• D. 29.4 g

Hint:

Use pi = (n/V)RT and then multiply the moles by the molar mass.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Osmotic pressure \(\pi = CRT = \dfrac{n}{V}RT\), where \(n = \dfrac{x}{M}\).

Step 2: Calculation
\(C = \dfrac{\pi}{RT} = \dfrac{0.8}{0.0821\times300} = 0.03248\) mol dm\(^{-3}\)
\(n = 0.03248\times3 = 0.09745\) mol

Step 3: Mass
\(x = n\times M = 0.09745\times196 = 19.1\) g

Step 4: Compare the options
This matches option (A).

Final Answer:
The mass of solute is 19.1 g. \[ \boxed{\text{(A) }19.1\ \text{g}} \]