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Standard XII

Chemistry

Kohlrausch Law

The specific ...

Question

The specific conductance of a saturated solution of AgCl at $25^{0}$ C after subtracting the specific conductance of water is 2.28 x $10^{- 4} S m^{- 1}$ . Calculate the solubility of AgCl in gram per $d m^{3}$ at this temperature. The molar ionic conductances of $A g^{+}$ and $C l^{-}$ ions are 73.3 x $10^{- 4}$ and 65.0 x $10^{- 4} S m^{2} m o l^{- 1}$ . (Ag: 108 and Cl: 35.5).

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Solution

$A g C l \to A g^{+} + C l^{-}$
$\land_{m}^{o} (A g C l) = λ^{o} A g^{+} + λ^{o} C l^{-}$
$\land_{m}^{o} = 73.3 \times 10^{- 4} + 65 \times 10^{- 4}$
$= 138.3 \times 10^{- 4} s m^{2} m o l^{- 1}$
Specific conductance $k = 2.28 \times 10^{- 4}$
Molarity $= \frac{2.28 \times 10^{- 4}}{138.3 \times 10^{- 4}} m o l / m^{3}$
$= 0.016485 \times 10^{3} m o l / d m^{3}$
$= 16.485 m o l / d m^{3}$
solubility = $16.485 \times (108 + 35.5) g / d m^{3}$
$= 2365 g / d m^{3}$

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AgCl→Ag++Cl−∧om(AgCl)=λoAg++λoCl−∧om=73.3×10−4+65×10−4 =138.3×10−4sm2mol−1Specific conductance k=2.28×10−4Molarity =2.28×10−4138.3×10−4mol/m3 =0.016485×103mol/dm3 =16.485mol/dm3solubility = 16.485×(108+35.5)g/dm3 =2365g/dm3