CBSE Class 12 Chemistry 2023: question 20(b)

Hints

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Step-by-step solution

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1. The radius is $r=0\cdot 5$ cm. Area of cross-section: $A=\pi r^2=3\cdot 14\times (0\cdot 5{)}^2=0\cdot 785{\text{cm}}^2$. The length is $l=50$ cm.

   ½ mark

2. Use $\kappa =\frac{l}{R\times A}=\frac{50\text{cm}}{5\cdot 55\times 10^3\Omega \times 0\cdot 785{\text{cm}}^2}$.

   1 mark

3. $\kappa =\frac{63\cdot 69}{5550}=1\cdot 147\times 10^{-2}{\text{S cm}}^{-1}$ (i.e. $11\cdot 47\times 10^{-3}$ S cm⁻¹).

   ½ mark

Answer: κ ≈ 1.147 × 10⁻² S cm⁻¹

Revise the chapter

• Chapter: Electrochemistry (notes, formulas, where marks are lost and every past question)