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प्रश्न
Conductivity of 2 × 10−3 M methanoic acid is 8 × 10−5 S cm−1. Calculate its molar conductivity and degree of dissociation if `∧_"m"^0` for methanoic acid, is 404 S cm2 mol−3.
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उत्तर
Given: Conductivity (K) = 8 × 10−5 S cm−1
Molarity (M) = 2 × 10−3 M
Formula:
Molar conductivity, `∧_"m"^"c" = ("K" xx 1000)/"C"`
Degree of dissociation, α = `∧_"m"^"c"/∧_"m"^0`
`∧_"m"^0 = ("K" xx 1000)/"C"` S cm2 mol−1
= `(8 xx 10^-5 xx 1000)/(2 xx 10^-3)`
= 40 S cm2 mol−1
`∝` = `(∧_"m"^"c")/(∧_"m"^0)`
= `40/404`
= 0.1
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संबंधित प्रश्न
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| Rahul set up an experiment to find the resistance of aqueous KCl solution for different concentrations at 298 K using a conductivity cell connected to a Wheatstone bridge. He fed the Wheatstone bridge with a.c. power in the audio frequency range 550 to 5000 cycles per second. Once the resistance was calculated from the null point, he also calculated the conductivity K and molar conductivity ∧m and recorded his readings in tabular form. |
| S. No. | Conc. (M) |
k S cm−1 | ∧m S cm2 mol−1 |
| 1. | 1.00 | 111.3 × 10−3 | 111.3 |
| 2. | 0.10 | 12.9 × 10−3 | 129.0 |
| 3. | 0.01 | 1.41 × 10−3 | 141.0 |
Answer the following questions:
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(c) If Rahul had used HCl instead of KCl then would you expect the ∧m values to be more or less than those per KCl for a given concentration? Justify. (2)
OR
(c) Amit a classmate of Rahul repeated the same experiment with CH3COOH solution instead of KCl solution. Give one point that would be similar and one that would be different in his observations as compared to Rahul. (2)
