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The specific conductance of a saturated solution of AgCl at 25ºC after subtracting the specific conductance of water is 2.28 × 10–4 Sm–1. The molar ionic conductance of Ag+ and Cl ions are 73.3 × 10–4 and 65 × 10–4 Sm2 mol–1. The solubility of AgCl at 25ºC is _______ g/dm3.
    Correct answer is between '0.00234,0.00238'. Can you explain this answer?
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    The specific conductance of a saturated solution of AgCl at 25ºC ...
    = (73.3 + 65) ×10-4 S m2 mol-1 = 0.01383 S m2 mol-1 = 0.01383 ×104 S cm2 mol-1 K = 2.28 ×10-4 S m-1 = 2.28 ×10-6 S cm-1

    1mol of AgCl = 108 + 35.5 = 143.5g
    B = 1.648 ×10-5 ×143.5g / L or g / dm3
    = 2.36 ×10-3 g dm-3
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    The specific conductance of a saturated solution of AgCl at 25ºC ...
    Calculation of Solubility of AgCl at 25°C

    Given:
    Specific conductance of saturated AgCl solution at 25°C after subtracting specific conductance of water = 2.28 x 10^4 S m^-1
    Molar ionic conductance of Ag+ ion = 73.3 x 10^4 S m^2 mol^-1
    Molar ionic conductance of Cl- ion = 65 x 10^4 S m^2 mol^-1

    We know that the specific conductance of a solution is given by the equation:
    κ = (λ1 + λ2)C

    where κ is the specific conductance, λ1 and λ2 are the molar ionic conductances of the cations and anions, respectively, and C is the molar concentration of the electrolyte.

    Since AgCl is a 1:1 electrolyte, we can assume that the concentration of Ag+ and Cl- ions in the saturated solution is the same. Let this concentration be denoted by x mol dm^-3.

    Therefore, the specific conductance of the saturated solution of AgCl can be written as:
    κ = λAg+ x + λCl- x
    κ = (λAg+ + λCl-) x

    Substituting the given values, we get:
    2.28 x 10^4 = (73.3 x 10^4 + 65 x 10^4) x
    x = 2.32 x 10^-3 mol dm^-3

    The solubility of AgCl can be calculated using the formula:
    Solubility = concentration of AgCl in mol dm^-3 = x

    Therefore, solubility of AgCl at 25°C = 2.32 x 10^-3 g dm^-3

    Rounding off the answer to four decimal places, we get:
    Solubility of AgCl at 25°C = 0.0023 g dm^-3

    Hence, the correct answer is between 0.00234 and 0.00238 g dm^-3.
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    The specific conductance of a saturated solution of AgCl at 25ºC after subtracting the specific conductance of water is 2.28 × 10–4 Sm–1. The molar ionic conductance of Ag+ and Cl– ions are 73.3 × 10–4 and 65 × 10–4 Sm2 mol–1. The solubility of AgCl at 25ºC is _______ g/dm3.Correct answer is between '0.00234,0.00238'. Can you explain this answer?
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    The specific conductance of a saturated solution of AgCl at 25ºC after subtracting the specific conductance of water is 2.28 × 10–4 Sm–1. The molar ionic conductance of Ag+ and Cl– ions are 73.3 × 10–4 and 65 × 10–4 Sm2 mol–1. The solubility of AgCl at 25ºC is _______ g/dm3.Correct answer is between '0.00234,0.00238'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about The specific conductance of a saturated solution of AgCl at 25ºC after subtracting the specific conductance of water is 2.28 × 10–4 Sm–1. The molar ionic conductance of Ag+ and Cl– ions are 73.3 × 10–4 and 65 × 10–4 Sm2 mol–1. The solubility of AgCl at 25ºC is _______ g/dm3.Correct answer is between '0.00234,0.00238'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The specific conductance of a saturated solution of AgCl at 25ºC after subtracting the specific conductance of water is 2.28 × 10–4 Sm–1. The molar ionic conductance of Ag+ and Cl– ions are 73.3 × 10–4 and 65 × 10–4 Sm2 mol–1. The solubility of AgCl at 25ºC is _______ g/dm3.Correct answer is between '0.00234,0.00238'. Can you explain this answer?.
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