Question

The specific conductivity of a saturated solution of $AgCl$ is $3.40 \times 10^{-6}$ ohm $^{-1}$ $cm ^{-1}$ at $25^{\circ} C .$ If $\lambda_{ Ag +}=62.3 ohm ^{-1} cm ^{2}$ $mol ^{-1} \& \lambda_{ cl }=67.7\, ohm ^{-1} cm ^{2} mol ^{-1}$, the solubility of $AgCl$ at $25^{\circ} C$ is :

• A. $2.6 \times 10^{-5} M$
• B. $4.5 \times 10^{-3} M$
• C. $3.6 \times 10^{-5} M$
• D. $3.6 \times 10^{-3} M$

Answer 1

Answer: (A)

$\lambda_{A g^{+}}=62.3 \,Scm ^{2} mol ^{-1}$,
$ \lambda_{cl^-}=67.7\, Scm ^{2} mole ^{-1}$
$K _{\text {Agcl }}=3.4 \times 10^{-6} Scm ^{-1}$
$\wedge_{ AgCl }^{\infty}=(62.3+67.5)$
$=\frac{1000 \times 3.4 \times 10^{-6}}{ S }$
$S =\frac{3.4 \times 10^{-3}}{(62.3+67.5)}$
$=2.6 \times 10^{-5} M$