The specific conductance of a saturated solution of AgCl at 298 K is found to be 1-386 -10 - 6Scm - 1- Calculate its solubility- - Ag - - 62-0Scm2mol - 1 and - Cl - - 76-3S cm2 mol - 1-

∧_m^∞_AgCl=λ_Ag^+^o+λ_Cl^⊝^o=62+76.3=138.3 ∧_m=KC× 1000 C=1.386× 10^ 6138.3× 1000=10^ 5 M ...

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Standard XII

Chemistry

Conductance and Conductivity

The specific ...

Question

The specific conductance of a saturated solution of $A g C l$ at 298 K is found to be $1.386 \times 10^{- 6} S c m^{- 1}$ . Calculate its solubility. $(λ_{A g^{+}}^{\circ} = 62.0 S c m^{2} m o l^{- 1} and λ_{C l^{-}}^{\circ} = 76.3 S c m^{2} m o l^{- 1})$ .

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Solution

${\land_{m}^{\infty}}_{A g C l} = λ_{A g^{+}}^{o} + λ_{C l^{⊝}}^{o} = 62 + 76.3 = 138.3$
$\land_{m} = \frac{K}{C} \times 1000$
$C = \frac{1.386 \times 10^{- 6}}{138.3} \times 1000 = 10^{- 5}$ $M$

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The specific conductance of a saturated solution of AgCl at 298 K is found to be 1.386×10−6Scm−1. Calculate its solubility.(λ∘Ag+=62.0Scm2mol−1andλ∘Cl−=76.3Scm2mol−1).

∧∞mAgCl=λoAg++λoCl⊝=62+76.3=138.3∧m=KC×1000C=1.386×10−6138.3×1000=10−5 M

The specific conductance of a saturated solution of $A g C l$ at 298 K is found to be $1.386 \times 10^{- 6} S c m^{- 1}$ . Calculate its solubility. $(λ_{A g^{+}}^{\circ} = 62.0 S c m^{2} m o l^{- 1} and λ_{C l^{-}}^{\circ} = 76.3 S c m^{2} m o l^{- 1})$ .

${\land_{m}^{\infty}}_{A g C l} = λ_{A g^{+}}^{o} + λ_{C l^{⊝}}^{o} = 62 + 76.3 = 138.3$
$\land_{m} = \frac{K}{C} \times 1000$
$C = \frac{1.386 \times 10^{- 6}}{138.3} \times 1000 = 10^{- 5}$ $M$