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Choose the Best Answer · Q19

Q.Conductivity of a saturated solution of a sparingly soluble salt AB (1:1 electrolyte) at 298 K is 1.85×10−5 S m−11.85\times10^{-5}\ \text{S m}^{-1}. Solubility product of the salt AB at 298 K is (Λmo)AB=14×10−3 S m2mol−1(\Lambda_m^{o})_{AB} = 14\times10^{-3}\ \text{S m}^2\text{mol}^{-1}.

(a) 5.7×10−125.7\times10^{-12}
(b) 1.32×10−121.32\times10^{-12}
(c) 7.5×10−127.5\times10^{-12}
(d) 1.74×10−121.74\times10^{-12}
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Step 1. For a very dilute saturated solution, molar conductivity is essentially equal to the limiting value: Λm≈Λmo\Lambda_m \approx \Lambda_m^{o}. Using SI units throughout (κ in S/m, Λm° in S m² mol⁻¹), the concentration follows from C(mol/m3)=κ/Λmo=1.85×10−514×10−3=1.321×10−3C(\text{mol/m}^3) = \kappa/\Lambda_m^{o} = \dfrac{1.85\times10^{-5}}{14\times10^{-3}} = 1.321\times10^{-3} mol/m³.

Step 2. Convert to mol/L by dividing by 1000 (since 1 m³ = 1000 L): C=1.321×10−3/1000=1.321×10−6C = 1.321\times10^{-3}/1000 = 1.321\times10^{-6} mol/L. …

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