हिंदी

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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प्रश्न

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

संख्यात्मक
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उत्तर

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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