For a saturated solution of AgCl at 250C- K-3-4-10-6 ohm-1 cm-1 and that of H2Ol used is 2-02-10-6 ohm-1- -0m for AgCl is 138 ohm-1cm2 mol-1-The solubility of AgCl in moles per liter is a - 10-5- Find a-

For a saturated solution of AgCl at 25^0C, λ ^0=1000× (K_sol K_H_2O)S 138=1000×(3.4 2.02)×10^ 6S S=1×10^ 5 ...

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Byju's Answer

Standard XII

Chemistry

Raoult's Law

For a saturat...

Question

For a saturated solution of AgCl at $25^{0} C, K = 3.4 \times 10^{- 6} o h m^{- 1} c m^{- 1}$ and that of $H_{2} O (l)$ used is $2.02 \times 10^{- 6} o h m^{- 1} . λ_{m}^{0}$ for AgCl is 138 $o h m^{- 1} c m^{2} m o l^{- 1} .$

The solubility of $A g C l$ in moles per liter is a $\times 10^{- 5}$ . Find a.

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Solution

For a saturated solution of AgCl at $25^{0} C,$
$λ^{0} = \frac{1000 \times (K_{s o l} - K_{H_{2} O})}{S}$
$138 = \frac{1000 \times (3.4 - 2.02) \times 10^{- 6}}{S}$
$S = 1 \times 10^{- 5}$

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