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Standard XII

Chemistry

Solubility Product

Limiting mola...

Question

Limiting molar conductivities of $M g^{2 +}$ and $C l^{-}$ ions are 106.0 S.cm $^{2}$ . mol $^{- 1}$ and
76.3 S.cm. $^{2}$ . mol $^{- 1}$ respectively. Calculate the limiting molar conductivity of $M g C l_{2}$ .

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Solution

$M g C l_{2} \to M g^{2 +} + 2 C l^{-}$

$Λ_{m}^{0} (M g C l_{2}) = λ_{m}^{0} (M g^{2 +}) + 2 λ_{m}^{0} (C l^{-})$

$Λ_{m}^{0} (M g C l_{2}) = 106 S \cdot c m^{2} / m o l + 2 \times 76.3 S \cdot c m^{2} / m o l = 258.6 S \cdot c m^{2} / m o l$

The limiting molar conductivity of $M g C l_{2}$ is $258.6 S \cdot c m^{2} / m o l$

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Limiting molar conductivities of Mg2+ and Cl− ions are 106.0 S.cm2. mol−1 and 76.3 S.cm.2. mol−1 respectively. Calculate the limiting molar conductivity of MgCl2.

MgCl2→Mg2++2Cl−Λ0m(MgCl2)=λ0m(Mg2+)+2λ0m(Cl−)Λ0m(MgCl2)=106S⋅cm2/mol+2×76.3S⋅cm2/mol=258.6S⋅cm2/mol The limiting molar conductivity of MgCl2 is 258.6S⋅cm2/mol

Limiting molar conductivities of $M g^{2 +}$ and $C l^{-}$ ions are 106.0 S.cm $^{2}$ . mol $^{- 1}$ and
76.3 S.cm. $^{2}$ . mol $^{- 1}$ respectively. Calculate the limiting molar conductivity of $M g C l_{2}$ .