CBSE Class 12 Chemistry 2024: question 22

Hints

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Step-by-step solution

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1. Cell constant from KCl: $G^{\ast }=\kappa \times R=1.35\times 10^{-2}\mathrm{S}\mathrm{c}{\mathrm{m}}^{-1}\times 100\Omega =1.35\mathrm{c}{\mathrm{m}}^{-1}$

   1 mark

2. Conductivity of AgNO₃: $\kappa =\frac{G^{\ast }}{R}=\frac{1.35}{90}=0.015\mathrm{S}\mathrm{c}{\mathrm{m}}^{-1}$

   1 mark

3. Molar conductivity: ${\Lambda }_m=\frac{\kappa \times 1000}{C}=\frac{0.015\times 1000}{0.02}=750\mathrm{S}\mathrm{c}{\mathrm{m}}^2\mathrm{mo}{\mathrm{l}}^{-1}$

   1 mark

Answer: κ = 0.015 S cm⁻¹; Λm = 750 S cm² mol⁻¹

Revise the chapter

• Chapter: Electrochemistry (notes, formulas, where marks are lost and every past question)
• Topic: Conductivity, Molar Conductivity and Kohlrausch's Law