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, $\alpha = \frac{\Lambda_{m}}{\Lambda_{m}^{o}} = \frac{22}{275} = 0.08$
Create a free account to view solution
View Solution FreeMore Electrochemistry Questions
ᐃG = ᐃH - TᐃS and ᐃG = ᐃH + T then is :...Which one of the following molecules has all the effect, namely inductive, mesomeric and hyperconjugative?...For a spontaneous reaction the ᐃG, equilibrium constant (K) and Eocell will be respectively...The current of 9.65 A flowing for 10 min deposits 3.0 g of a metal. The equivalent weight of the metal is...MnO4- + 8H+ + 5e- → Mn2+ + 4H2O, If H+ concentration is decreased from 1 M to 10-4 M at 25oC, where as concentrati...