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NEET CHEMISTRYElectrochemistryEasy

Question

The molar conductivity of a solution of AgNO3\text{AgNO}_3 at 298 K298 \text{ K} (considering the given data: concentration =0.5 mol/dm3= 0.5 \text{ mol/dm}^3, electrolytic conductivity =5.76×103 S cm1= 5.76 \times 10^{-3} \text{ S cm}^{-1}) is:

A

2.88 S cm2 mol12.88 \text{ S cm}^2 \text{ mol}^{-1}

B

11.52 S cm2 mol111.52 \text{ S cm}^2 \text{ mol}^{-1}

C

0.086 S cm2 mol10.086 \text{ S cm}^2 \text{ mol}^{-1}

D

28.8 S cm2 mol128.8 \text{ S cm}^2 \text{ mol}^{-1}

Step-by-Step Solution

The molar conductivity (Λm\Lambda_m) is calculated using the formula:

Λm=κ×1000C\Lambda_m = \frac{\kappa \times 1000}{C}

Where: κ\kappa (electrolytic conductivity) =5.76×103 S cm1= 5.76 \times 10^{-3} \text{ S cm}^{-1} CC (concentration/molarity) =0.5 mol/dm3=0.5 mol L1= 0.5 \text{ mol/dm}^3 = 0.5 \text{ mol L}^{-1}

Substituting the values into the formula: Λm=5.76×103×10000.5\Lambda_m = \frac{5.76 \times 10^{-3} \times 1000}{0.5} Λm=5.760.5=11.52 S cm2 mol1\Lambda_m = \frac{5.76}{0.5} = 11.52 \text{ S cm}^2 \text{ mol}^{-1}.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Electrochemistry. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYElectrochemistryconductivitysolutiontextagnoconsideringconcentration

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