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Chemistry
Limiting molar conductivity of NH 4 OH [i.e., Lambdam°( NH 4 OH ) ] is equal to:
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Q. Limiting molar conductivity of $NH _{4} OH$ [i.e., $\Lambda_{m}^{\circ}\left( NH _{4} OH \right)$ ] is equal to:
AIPMT
AIPMT 2012
Electrochemistry
A
$\Lambda_{m}^{\circ}\left(N H_{4} C l\right)+\Lambda_{m}^{\circ}\left(N a_{4} C l\right)-\Lambda_{m}^{\circ}(N a O H)$
7%
B
$\Lambda_{m}^{\circ}(N a O H)+\Lambda_{m}^{\circ}(N a C l)-\Lambda_{m}^{\circ}\left(N H_{4} C l\right)$
34%
C
$\Lambda_{m}^{\circ}\left(N_{4} O H\right)+\Lambda_{m}^{\circ}\left(N H_{4} C l\right)-\Lambda_{m}^{\circ}(H C l)$
20%
D
$\Lambda_{m}^{\circ}\left( NH _{4} Cl \right)+\Lambda_{m}^{\circ}( NaOH )-\Lambda_{m}^{\circ}( NaCl )$
38%
Solution:
According to Kohlrausch law of independent migration of ions:
$\Lambda_{m}^{\circ}\left( NH _{4} OH \right)=\Lambda_{m}^{\circ}\left( NH _{4}^{+}\right)+\Lambda_{m}^{\circ}\left( OH ^{-}\right)$
$=\Lambda_{m}^{\circ}\left( NH _{4}^{+}\right)+\Lambda_{m}^{\circ}\left( Cl ^{-}\right)+\Lambda_{m}^{\circ}\left( OH ^{-}\right)$
$=\Lambda_{m}^{\circ}\left( NH _{4}^{+}\right)-\Lambda_{m}^{\circ}\left( Cl ^{-}\right)-\Lambda_{m}^{\circ}\left( Na ^{+}\right)$
$=\Lambda_{m}^{\circ}\left( NH _{4} Cl \right)+\Lambda_{m}^{\circ}( NaOH )-\Lambda_{m}^{\circ}( NaCl )$