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Question
Calculate molar conductivities at zero concentration for CaCl2 and NaCl.
Given: molar ionic conductivities of Ca2+, Cl–, Na+ ions are respectively, 104, 76.4, 50.1 Ω–1 cm–2 mol–1
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Solution
Given:
`lambda_("Ca"^(2+))^0` = 104 Ω–1 cm2 mol–1,
`lambda_("Cl"^(-))^0` = 76.4 Ω–1 cm2 mol–1,
`lambda_("Na"^(+))^0` = 50.1 Ω–1 cm2 mol–1,
To find: Molar conductivity at zero concentration for CaCl2 and NaCl:
∧0 (CaCl2) and ∧0 (NaCl)
Formula: ∧0 = `"n"_+ lambda_+^0 + "n"_(-)lambda_(-)^0`
Calculation: According to Kohlrausch law,
∧0(CaCl2) = `lambda_("Ca"^(2+))^0` + `2lambda_("Cl"^(-))^0`
= 104 Ω–1 cm2 mol–1 + 2 × 76.4 Ω–1 cm2 mol–1
= 256.8 Ω–1 cm2 mol–1
∧0(NaCl) = `lambda_("Na"^(+))^0` + `lambda_("Cl"^(-))^0`
= 50.1 Ω–1 cm2 mol–1 + 76.4 Ω–1 cm2 mol–1
= 126.5 Ω–1 cm2 mol–1
Molar conductivity at zero concentration of CaCl2 is 256.8 Ω–1 cm2 mol–1
Molar conductivity at zero concentration of NaCl is 126.5 Ω–1 cm2 mol–1
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