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प्रश्न
Calculate the degree of dissociation (α) of acetic acid if its molar conductivity (Λm) is 39.05 S cm2 mol−1.
(Given \[\ce{\lambda^{\circ}_{(H^+)}}\] = 349.6 S cm2 mol−1 and \[\ce{\lambda^{\circ}_{(CH_3COO^-)}}\] = 40.95 S cm2 mol−1)
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उत्तर
The degree of dissociation of a weak electrolyte is given by the following relation:
α = `Lambda_m/Lambda^circ` ... (1)
Where,
α = Degree of dissociation of the electrolyte
Λm = Molar conductivity
Λ° = Molar conductivity at infinite dilution
We are given that the ionic molar conductivity at infinite dilution of acetate and hydrogen ions are 349.8 and 40.9 S cm2 mol−1, respectively.
Hence, the limiting molar conductivity of acetic acid would be written as:
\[\ce{\Lambda^{\circ}_{\text{acetic acid}} = \lambda^{\circ}_{(H^+)} + \lambda^{\circ}_{(CH_3COO^-)}}\]
= 40.9 + 349.8
= 390.7 cm2 mol−1
Now, since we are given that the molar conductivity of acetic acid is:
Λm = 39.05 S cm2 mol−1
Therefore, substituting the values in equation (1), we get:
Degree of dissociation of acetic acid (α) = `Lambda_m/Lambda^circ`
= \[\ce{\frac{39.05}{390.7}}\]
= 0.0999
≅ 0.1
Hence, the degree of dissociation is 0.1.
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संबंधित प्रश्न
Define “Molar conductivity”.
The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below:
| 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 |
Calculate ∧m for all concentrations and draw a plot between ∧m and c1/2. Find the value of `Lambda_m^0`.
Write mathematical expression of molar conductivity of the given solution at infinite dilution.
A steady current of 2 amperes was passed through two electrolytic cells X and Y connected in series containing electrolytes FeSO4and ZnSO4 until 2.8g of Fe deposited at the cathode of cell X. How long did the current flow? Calculate the mass of Zn deposited at the cathode of cell Y.
(Molar mass: Fe=56g mol-1,Zn=65.3g mol-1,1F=96500C mol-1)
Molar conductivity denoted by the symbol Λm is related to the conductivity of the solution by the equation (k is the conductivity and c is the concentration).
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Choose the right option for your answer.
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The following questions are case-based questions. Read the passage carefully and answer the questions that follow:
| 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:
(a) Why does conductivity decrease with dilution? (1)
(b) If `∧_"m"^0` of KCl is 150.0 S cm2 mol−1, calculate the degree of dissociation of 0.01 M KCI. (1)
(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)
The solution of two electrolytes A and B are diluted. ^m of B increases 1.5 times while that of A increases 25 times. Which of the two is a strong electrolyte? Give a reason.
