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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 =5.76×10−3 S cm−1= 5.76 \times 10^{-3} \text{ S cm}^{-1}) is:

A

2.88 S cm2 mol−12.88 \text{ S cm}^2 \text{ mol}^{-1}

B

11.52 S cm2 mol−111.52 \text{ S cm}^2 \text{ mol}^{-1}

C

0.086 S cm2 mol−10.086 \text{ S cm}^2 \text{ mol}^{-1}

D

28.8 S cm2 mol−128.8 \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) =5.76×10−3 S cm−1= 5.76 \times 10^{-3} \text{ S cm}^{-1} CC (concentration/molarity) =0.5 mol/dm3=0.5 mol L−1= 0.5 \text{ mol/dm}^3 = 0.5 \text{ mol L}^{-1}

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

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