A solution of 0.1 M $C H_{3} C O O H$ is placed between parallel electrodes of cross-section
area $4 c m^{2}$ , separated by 2 cm. For this solution, resistance measured is $100 Ω$ . Calculate elevation in boiling point of the 0.1 M $C H_{3} C O O H$ solution, using following information $K_{b} = 0.5 K k g / m o l; Λ_{m}^{\infty} (H^{+}) = 300 S c m^{2} m o l e^{- 1};$
$Λ_{m}^{\infty} (C H_{3} C O O^{-}) = 100 S c m^{2} m o l e^{- 1}$

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answer is 0.06.

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Detailed Solution

$\frac{l}{A} = \frac{2}{4}; R = ρ . \frac{1}{2} o r ρ = 200 Ω κ = \frac{1}{200} Ω^{- 1} c m^{- 1} \begin{array}{l} Λ = \frac{1}{200} \times \frac{1000}{0.1} = 50 \\ Λ^{0} = 300 + 100 = 400 \\ α = \frac{Λ}{Λ^{0}} = \frac{50}{400} = \frac{1}{8} \\ Λ T_{b} = i \times K_{b} \times m \\ = (1 + \frac{1}{8}) \times 0.5 \times 0.0.05625 \end{array}$