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At 300K molar conductivity of solution A is 350 units, and at infinite dilution the molar conductivity of the same sample is 480 unit. Predict the percentage dissociation of the electrolyte.
- a)73.0%
- b)37.0%
- c)63.0%
- d)137.0%
Correct answer is option 'A'. Can you explain this answer?
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At 300K molar conductivity of solution A is 350 units, and at infinite...
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At 300K molar conductivity of solution A is 350 units, and at infinite...
Given Data:
Molar conductivity of solution A at 300K = 350 units
Molar conductivity of solution A at infinite dilution = 480 units
Calculating Percentage Dissociation:
The molar conductivity of an electrolyte at infinite dilution is the sum of the molar conductivities of its constituent ions. Therefore, we can write:
Molar conductivity at infinite dilution = λcation + λanion
Let's assume that the electrolyte dissociates into cation (A+) and anion (B-)
At infinite dilution, the molar conductivity of the electrolyte is equal to the sum of the molar conductivities of its constituent ions:
480 units = λA+ + λB-
Now, according to Kohlrausch's law of independent migration of ions, the molar conductivity (λ) of an electrolyte at a given concentration is related to its molar conductivity at infinite dilution (λ0) by the equation:
λ = λ0 - A√C
where λ is the molar conductivity at concentration C, λ0 is the molar conductivity at infinite dilution, and A is a constant.
Let's consider the molar conductivity of solution A at 300K:
350 units = λ0 - A√C
Now, we can calculate the percentage dissociation (α) using the formula:
α = (λ0 - λ) / λ0 * 100
Calculating α:
From the given data, we know that:
λ = 350 units
λ0 = 480 units
Substituting these values in the formula for α:
α = (480 - 350) / 480 * 100
α = 130 / 480 * 100
α ≈ 27.08%
Conclusion:
The percentage dissociation of the electrolyte in solution A is approximately 27.08%. Therefore, none of the given options (a, b, c, d) is the correct answer.
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