ElectrochemistryHard
Question
The conductivity of a saturated solution of AgCl at 298 K was found to be 3.40 × 10−6 Ω−1 cm−1; the conductivity of water used to make up the solution was 1.60 × 10−6 Ω−1 cm−1. Determine the solubility of AgCl in water in mole per litre at 298 K. The equivalent conductivity of AgCl at infinite dilution is 150.0 Ω−1 cm−2 eq−1.
Options
A.1.44 × 10−10
B.1.2 × 10−5
C.3.33 × 10−5
D.1.2 × 10−8
Solution
$\Lambda_{eq} = \frac{\kappa}{C} \Rightarrow 150 \times \frac{3.4 \times 10^{- 6} - 1.6 \times 10^{- 6}}{5}$
$\therefore S = 1.2 \times 10^{- 8}\text{ mol c}\text{m}^{- 3} = 1.2 \times 10^{- 5}\text{ M}$
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