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प्रश्न
The conductivity of 0.02 M AgNO3 at 25°C is 2.428 × 10−3 Ω−1 cm−1. What is its molar conductivity?
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उत्तर
Given: C = 0.02 M
k = 2.428 × 10−3Ω−1 cm−1
To find: Molar conductivity = ?
Formula: Molar conductivity = `(1000k)/C`
= `(1000 xx 2.428 xx 10^-3)/0.02`
= `(2425.0 xx 10^-3)/0.02`
= `2.425/0.02`
= 121.25 Ω−1 cm−1 mol−1
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संबंधित प्रश्न
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| Concentration/M | 0.001 | 0.010 | 0.020 | 0.050 | 0.100 |
| 102 × κ/S m−1 | 1.237 | 11.85 | 23.15 | 55.53 | 106.74 |
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In the plot of molar conductivity (∧m) vs square root of concentration (c1/2), following curves are obtained for two electrolytes A and B:
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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 |
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OR
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