The molar conductivity of a 0.5 mol/dm30.5 \text{ mol/dm}^30.5 mol/dm3 solution of AgNO3\text{AgNO}_3AgNO3 with electrolytic conductivity of 5.76×10−3 S cm−15.76 \times 10^{-3} \text{ S cm}^{-1}5.76×10−3 S cm−1 at 298 K298 \text{ K}298 K is:
11.5 S cm2/mol11.5 \text{ S cm}^2/\text{mol}11.5 S cm2/mol
21.5 S cm2/mol21.5 \text{ S cm}^2/\text{mol}21.5 S cm2/mol
31.5 S cm2/mol31.5 \text{ S cm}^2/\text{mol}31.5 S cm2/mol
41.5 S cm2/mol41.5 \text{ S cm}^2/\text{mol}41.5 S cm2/mol
Given: Concentration, C=0.5 mol/dm3=0.5 mol/LC = 0.5 \text{ mol/dm}^3 = 0.5 \text{ mol/L}C=0.5 mol/dm3=0.5 mol/L Conductivity, κ=5.76×10−3 S cm−1\kappa = 5.76 \times 10^{-3} \text{ S cm}^{-1}κ=5.76×10−3 S cm−1
The formula for molar conductivity (Λm\Lambda_mΛm) is: Λm=κ×1000C\Lambda_m = \frac{\kappa \times 1000}{C}Λm=Cκ×1000
Substituting the given values: Λm=5.76×10−3×10000.5=5.760.5=11.52 S cm2mol−1≈11.5 S cm2/mol\Lambda_m = \frac{5.76 \times 10^{-3} \times 1000}{0.5} = \frac{5.76}{0.5} = 11.52 \text{ S cm}^2\text{mol}^{-1} \approx 11.5 \text{ S cm}^2/\text{mol}Λm=0.55.76×10−3×1000=0.55.76=11.52 S cm2mol−1≈11.5 S cm2/mol
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