ElectrochemistryHard
Question
The molar conductivity of NH4Cl, OH− and Cl− at infinite dilution is 150, 200 and 75 Ω−1 cm2 mol−1, respectively. If the molar conductivity of a 0.01 M−NH4OH solution is 22 Ω−1 cm2 mol−1, then its degree of dissociation is
Options
A.0.146
B.0.063
C.0.080
D.0.293
Solution
$\Lambda_{m\left( NH_{4}OH \right)}^{o} = \Lambda_{m\left( NH_{4}Cl \right)}^{o} - \Lambda_{m\left( Cl^{-} \right)}^{o} + \Lambda_{m\left( OH^{-} \right)}^{o}$
= 150 – 75 + 200 = 275 ohm–1 cm2 mol–1
Now, $\alpha = \frac{\Lambda_{m}}{\Lambda_{m}^{o}} = \frac{22}{275} = 0.08$
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