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Question #44762

Difficulty: 8/10 | 🟢 Free

The Question

A $0.05 \text{ M}$ aqueous solution of a weak monoprotic acid $HA$ is placed in a conductivity cell with a cell constant of $0.5 \text{ cm}^{-1}$ at $300 \text{ K}$ . The measured resistance of the solution is $250 \, \Omega$ . Given the limiting molar conductivities $\lambda^\circ(H^+) = 350 \text{ S cm}^2 \text{ mol}^{-1}$ and $\lambda^\circ(A^-) = 50 \text{ S cm}^2 \text{ mol}^{-1}$ , calculate the osmotic pressure of this solution at $300 \text{ K}$ .

(Assume $RT = 24.6 \text{ L atm mol}^{-1}$ )

Options

A.
1.353 atm
B.
1.230 atm
C.
2.460 atm
D.
1.107 atm

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Numerical_Precision

Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.

Concepts Behind This Question

Molar Conductivity Trend|Van’t Hoff Factor (i) Application

Why This Question Is Tricky

Common Mistakes

This problem bridges Electrochemistry and Colligative Properties.

The 'Hidden Variable' tested is the concentration-dependent Van't Hoff factor ( $i$ ) for weak electrolytes.

Students often fall into the trap of using the stoichiometric factor ( $i=2$ ) assuming complete dissociation, or ignoring it entirely ( $i=1$ ).

The correct approach requires calculating the degree of dissociation ( $\alpha$ ) via the Arrhenius conductivity ratio ( $\Lambda_m / \Lambda_m^\circ$ ) and applying it to find the true effective particle concentration ( $i = 1 + \alpha$ ).

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Electrochemistry - Class 12
Concept	Molar Conductivity Trend
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free