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

Question

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 (A) 39% (B) 3.9% (C) 0.35 % (D) 0.039%

Tardigrade Admin · Accepted Answer

( Lambda0 C6H5COOH) = λ0C6H5COO- + λ0 H+ = 42+ 288.42 = 330.42 a = ( Lambda°m/ Lambdam°) = (12.8/330.42) = 0.039 = 3.9 %

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

39%

3.9%

0.35 %

0.039%

$(Λ^{0} C_{6} H_{5} COO H) = λ_{C_{6} H_{5} CO O^{-}}^{0} + λ_{H^{+}}^{0}$
$= 42 + 288.42 = 330.42$
$a = \frac{Λ _{m}^{\circ}}{Λ _{m^{\circ}}} = \frac{12.8}{330.42}$
$= 0.039 = 3.9%$

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

39%

3.9%

0.35 %

0.039%

(Λ0C6​H5​COOH)=λC6​H5​COO−0​+λH+0​ =42+288.42=330.42 a=Λm∘​Λm∘​​=330.4212.8​ =0.039=3.9%

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

$(Λ^{0} C_{6} H_{5} COO H) = λ_{C_{6} H_{5} CO O^{-}}^{0} + λ_{H^{+}}^{0}$
$= 42 + 288.42 = 330.42$
$a = \frac{Λ _{m}^{\circ}}{Λ _{m^{\circ}}} = \frac{12.8}{330.42}$
$= 0.039 = 3.9%$