Question

HA is a weak electrolyte. At ${25}^{0}C$, the degree of dissociation of 0.5 mol $L^{-1}$ HA is 0.1. What is its molar conductivity ( in $Sc{m}^{2}mo{l}^{-1}$)?
(The limiting molar conductivity of HA is $390Sc{m}^{2}mo{l}^{-1}$)

• A. 39
• B. 195
• C. 3.9
• D. 390

Answer 1

Answer: (A)

39

Degree of dissociation $=\frac{\text{Molar conductivity of O.S.M electrolyte}}{\text{Limiting molar conductivity}}$

$\therefore \alpha =0.1=\frac{\text{molar conductivity}}{390}$

$\therefore$ Molar conductivity $=39Sc{m}^{2}mo{l}^{-1}$