SATHEE: Electrochemistry - Result Question 1

Electrochemistry - Result Question 1

####1. Following limiting molar conductivities are given as

$λ_{m}^{\circ} (H_{2} S O_{4}) = x S c m^{2} m o l^{- 1}$

[NEET Odisha 2019]

$λ_{m}^{\circ} (K_{2} S O_{4}) = y S c m^{2} m o l^{- 1}$

$λ_{m}^{\circ} (C H_{3} C O O K) = z S c m^{2} m o l^{- 1}$

$λ_{m}^{\circ}$ (in $S c m^{2} m o l^{- 1}$ ) for $C H_{3} C O O H$ will be

(a) $\frac{(x - y)}{2} + z$

(b) $x - y + 2 z$

(d) $x - y + z$

Show Answer

Solution:

(a) According to Kohlrausch’s law

$λ_{m}^{\circ} (A B) = λ_{m}^{\circ} (A^{+}) + λ_{m}^{\circ} (B -)$

So, $λ_{m}^{\circ} (C H_{3} C O O H) = λ_{m}^{\circ} (C H_{3} C O O^{-}) + λ_{m}^{\circ} (H^{+})$ So $λ_{m}^{\circ} (C H_{3} C O O H)$

$= λ_{m}^{\circ} (C H_{3} C O O K) + \frac{1}{2} λ_{m}^{\circ} (H_{2} S O_{4}) - \frac{1}{2} λ_{m}^{\circ} (K_{2} S O_{4})$

$= z + \frac{x}{2} - \frac{y}{2} = z + (\frac{x - y}{2})$

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Electrochemistry - Result Question 1

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