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Question 72 of 73

Q.The molar conductivity of a 0.5 mol dm−30.5\ mol\ dm^{-3} solution of AgNO3AgNO_3 with specific conductance 5.76×10−3 S cm−15.76 \times 10^{-3}\,S\,cm^{-1} at 298 K is :

(a) 0.086 S cm2 mol−10.086\ S\,cm^2\,mol^{-1}
(b) 2.88 S cm2 mol−12.88\ S\,cm^2\,mol^{-1}
(c) 28.8 S cm2 mol−128.8\ S\,cm^2\,mol^{-1}
(d) 11.52 S cm2 mol−111.52\ S\,cm^2\,mol^{-1}
Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2026MCQ· 1mImportance★★★★★
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Molar conductivity is obtained from the specific conductance by dividing by the molar concentration (after converting volume units from litres to cubic centimetres via the factor 1000); substituting the given values directly gives the answer.

Formula: Λm=κ×1000C\Lambda_m = \dfrac{\kappa \times 1000}{C} where κ\kappa is the specific conductance (in S cm−1S\,cm^{-1}) and CC is the molar concentration (in mol dm−3=mol L−1mol\,dm^{-3} = mol\,L^{-1}), giving Λm\Lambda_m in S cm2 mol−1S\,cm^2\,mol^{-1}.

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