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Practice question

Question

What is the ll constant if the resistance of a conductivity ll with 0.01 M KCl solution is 200 Ω and its conductivity is 0.14 × 10⁻² S cm⁻¹?

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Explanation

Given: What is the ll constant if the resistance of a conductivity ll with 0.01 M KCl solution is 200 Ω and its conductivity is 0.14 × 10⁻² S cm⁻¹? These values define the system as per NCERT data. Formula: Cell constant, G^* = kappa × R = 0.14 × 10⁻² × 200 = 0.28 cm^{-1 .. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Substituting values like 1.2 × 10⁻⁵, 236 J kg⁻¹ K⁻¹, CH₃CH₂NH₂ etc. into the formula and simplifying step by step. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

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