Conductivity of a saturated solution of Cu2-FeCN6- after subtracting the conductivity of water is 1-28 -10-5-1cm-1-Calculate value of solubillity of Cu-FeCN6- -mxCuSO4-260 S cm2 mol-1 - -mxK2SO4-300 S cm2 mol-1- -mxK4FeCN6-720 S cm2 mol-1-Report your answer as solubility - 105 -

K _ 4 Fe( CN ) #160; _ 6 +2Cu SO _ 4 → Cu _ 2 [ Fe( CN ) #160; _ 6 ] +2 H _ 2 So _ 4 ∴ λ #160; _ m Cu _ 2 [ Fe( CN ) #160; _ 6 ] =λ #160; ...

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Byju's Answer

Standard XII

Chemistry

Kohlrausch Law

Conductivity ...

Question

Conductivity of a saturated solution of $C u_{2} [F e (C N)_{6}]$ after subtracting the conductivity of water is $1.28 \times 10^{- 5} Ω^{- 1} c m^{- 1}$ .
Calculate value of solubillity of $C u [F e (C N)_{6}]$ .
$[Λ_{m}^{x} (C u S O_{4}) = 260 S c m^{2} m o l^{- 1}$ ,
$Λ_{m}^{x} (K_{2} S O_{4}) = 300 S c m^{2} m o l^{- 1}$ ,
$Λ_{m}^{x} (K_{4} F e (C N)_{6}) = 720 S c m^{2} m o l^{- 1}]$
Report your answer as (solubility) $\times (10^{5})$ .

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Solution

$K_{4} F e {(C N)}_{6} + 2 C u {S O}_{4} \to {C u}_{2} [F e {(C N)}_{6}] + 2 H_{2} {S o}_{4}$
$∴ λ_{m} {C u}_{2} [F e {(C N)}_{6}] = λ_{m} K_{4} F e {(C N)}_{6} + 2 λ_{m} C u {S O}_{4} - 2 λ_{m} K_{2} {S O}_{4}$
$λ_{m} = 720 + 2 \times 260 - 2 \times 300$
$= 640 {s c m}^{2} {m o l}^{- 1}$

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