1. Tardigrade
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  3. Chemistry
  4. If equivalent conductance of 1 M benzoic acid is 12.8 ohm-1 cm2 and if the conductance of benzoate ion and H+ ion are 42 and 288.42 ohm-1 respectively, its degree of dissociation is

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Q. If equivalent conductance of 1 M benzoic acid is $12.8$ $ohm^{-1} cm^2$ and if the conductance of benzoate ion and $H^+$ ion are 42 and 288.42 $ohm^{-1}$ respectively, its degree of dissociation is

Electrochemistry

Solution:

$\left(\Lambda^{0} C_{6}H_{5}COOH\right) = \lambda^{0}_{C_6H_5COO^{-}} + \lambda^{0} _{H^{+}} $
$= 42+ 288.42 = 330.42$
$ a = \frac{\Lambda^{\circ}_{m}}{\Lambda_{m^{\circ}}} = \frac{12.8}{330.42}$
$ = 0.039 = 3.9 \%$