Electrochemistry - Result Question 104
52. The conductance of a $0.0015 $ $M$ aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinised Pt electrodes. The distance between the electrodes is $120 $ $cm$ with an area of cross section of $1 $ $cm^2$ . The conductance of this solution was found to be $5 \times 10^{-7}$ $S$. The $pH$ of the solution is $4$ . The value of limiting molar conductivity $\left(\Lambda _m^{\circ}\right)$ of this weak monobasic acid in aqueous solution is $Z \times 10^{2} $ $S$ $ cm^{-1} mol^{-1}$. The value of $Z$ is
(2017 Adv.)
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Answer:
Correct Answer: 52. $(6 \times 10^{2} $ $S$ $ cm^{-1} mol^{-1})$
Solution:
- $pH=C \alpha=10^{-4}$
$ \Rightarrow \alpha=\frac{10^{-4}}{0.0015} $
Also, conductance $(G)=\kappa\left(\frac{A}{l}\right)$
$\Rightarrow \kappa =G\left(\frac{l}{A}\right)=5 \times 10^{-7} \times \frac{120}{1}=6 \times 10^{-5} $
$\Rightarrow \Lambda^{c} =\frac{\kappa \times 1000}{C} $
$ =\frac{6 \times 10^{-5} \times 1000}{0.0015} $
$\Rightarrow \Lambda^{\infty} =\frac{\Lambda^{c}}{\alpha}=\frac{6 \times 10^{-5} \times 1000}{0.0015} \times \frac{0.0015}{10^{-4}} $
$ =600=6 \times 10^{2} $ $S$ $ cm^{-1} mol^{-1}$
BETA
