SATHEE: Electrochemistry - Result Question 12

Electrochemistry - Result Question 12

####13. Equivalent conductances of $N a C l, H C l$ and $C H_{3} C O O N a$ at infinite dilution are 126.45,426.16 and $91 o h m^{- 1} c m^{2}$ respectively. The equivalent conductance of $C H_{3} C O O H$ at infinite dilution would be

[1997]

(a) $101.38 o h m^{- 1} c m^{2}$

(b) $253.62 o h m^{- 1} c m^{2}$

(d) $678.90 o h m^{- 1} c m^{2}$

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Solution:

$λ_{N a^{+}}^{\circ} + λ_{C l^{-}}^{\circ} = 126.45$

$λ_{H^{+}}^{\circ} + λ_{C l^{-}}^{\circ} = 426.16$

$λ_{C H_{3} O O^{-}}^{\circ} + λ_{N a^{+}}^{\circ} = 91$

on adding (2) and (3) then subtract (1) from it

$λ_{C H_{3} C O O^{-}}^{\circ} + λ_{H^{+}}^{\circ} = 517.16 - 126.45$

$\graveU_{(C H_{3} C O O H)}^{\circ} = 390.71 o h m^{- 1} c m^{2}$

Electrochemistry - Result Question 12

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Electrochemistry - Result Question 12

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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