CBSE Class 12 Chemistry 2024: question 21

Hints

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Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

1. Find the cell constant from the 0.2 M solution:
   $G^{\ast }=\kappa \times R=0.0248{\text{S cm}}^{-1}\times 200\Omega =4.96{\text{cm}}^{-1}$

   ½ mark

2. Conductivity of the 0.05 M solution:
   $\kappa =\frac{G^{\ast }}{R}=\frac{4.96}{620}=0.008{\text{S cm}}^{-1}$

   ½ mark

3. Molar conductivity formula:
   ${\Lambda }_m=\frac{\kappa \times 1000}{c}$

   ½ mark

4. Substitute:
   ${\Lambda }_m=\frac{0.008\times 1000}{0.05}=160{\text{S cm}}^2{\text{ mol}}^{-1}$

   ½ mark

Answer: $\kappa =0.008$ S cm${}^{-1}$; ${\Lambda }_m=160$ S cm${}^2$ 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