What will be the molar conductivity of AL3+ ions at infinite dilution if molar conductivity of Al2(SO4)3 is 858 S cm2 mol-1 and ionic conductance of SO2-4 is 160 S cm2 mol-1 at infinite dilution?

Question

What will be the molar conductivity of AL3+ ions at infinite dilution if molar conductivity of Al2(SO4)3 is 858 S cm2 mol-1 and ionic conductance of SO2-4 is 160 S cm2 mol-1 at infinite dilution? (A) 189 S cm2 mol-1 (B) 698 S cm2 mol-1 (C) 1018 S cm2 mol-1 (D) 429 S cm2 mol-1

Tardigrade Admin · Accepted Answer

Lambda°Al2(SO4)3=2 Lambda°Al3++3 Lambda°So2-4 Lambda°Al3+=( Lambda°Al2(SO4)3-3 Lambda°SO2-4/2) =(858-(3×160)/2) =189 S cm2 mol-1

Q. What will be the molar conductivity of $A L^{3 +}$ ions at infinite dilution if molar conductivity of $A l_{2} (S O_{4})_{3}$ is $858 S c m^{2} m o l^{- 1}$ and ionic conductance of $S O_{4}^{2 -}$ is $160 S c m^{2} m o l^{- 1}$ at infinite dilution?

$189 S c m^{2} m o l^{- 1}$

$698 S c m^{2} m o l^{- 1}$

$1018 S c m^{2} m o l^{- 1}$

$429 S c m^{2} m o l^{- 1}$

$Λ_{A l_{2} (S O_{4})_{3}}^{\circ} = 2 Λ_{A l^{3 +}}^{\circ} + 3 Λ_{S o_{4}^{2 -}}^{\circ}$
$Λ_{A l^{3 +}}^{\circ} = \frac{Λ _{A l_{2}}^{\circ} ( S O _{4} ) _{3} - 3 Λ _{S O_{4}^{2 -}}^{\circ}}{2}$
$= \frac{858 - ( 3 \times 160 )}{2}$
$= 189 S c m^{2} m o l^{- 1}$

Q. What will be the molar conductivity of AL3+ ions at infinite dilution if molar conductivity of Al2​(SO4​)3​ is 858Scm2mol−1 and ionic conductance of SO42−​ is 160Scm2mol−1 at infinite dilution?

189Scm2mol−1

698Scm2mol−1

1018Scm2mol−1

429Scm2mol−1