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प्रश्न
Discuss the variation of conductivity and molar conductivity with concentration.
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उत्तर
Both conductivity and molar conductivity change with the change in concentration of electrolyte. Conductivity always decreases with decreasing the concentration of both weak and strong electrolytes. This can be explained by the fact that with dilution, the number of ions carrying electric current per unit volume decreases. The conductivity of a solution at any concentration is the conductivity of a unit volume of the solution placed between two platinum electrodes having a unit cross-sectional area and situated at a unit distance from each other.
This is clear from the following equation:
C = `(kappa A)/l` = κ ...(Both A and I are in appropriate units m or cm)
The molar conductivity of a solution at a given concentration is the conductivity of volume (V) of the solution containing one mole of electrolyte dissolved in it and placed between two electrodes of cross-sectional area (A), located at a unit distance from each other. So,
Plot of c1/2 versus molar conductivity for potassium chloride (a strong electrolyte) in aqueous solution.
∧m = `(kappa A)/l` = κ
Since l = 1 and A = V (volume in which one mole of electrolyte is dissolved.)
∧m = κ V
Molar conductivity increases with decreasing concentration. This is because the total volume (V) in which one mole of electrolyte is present also increases. It has been found that on dilution of the solution, the increase in volume is much more than the decrease in κ.
Strong Electrolytes: For strong electrolytes, the value of ∧m increases gradually with dilution, and it can be represented by the following equation:
∧m = `∧_m^0 - Ac^(1//2)`
It can be seen that if ∧m is plotted against c1/2, we get a straight line with intercept A and slope equal to ‘A’. The value of constant ‘A’ at a given solvent and temperature depends on the type of electrolyte, i.e., on the charges of the cation and anion produced on dissociation of the electrolyte in solution. Thus, NaCl, CaCl2, MgSO4 are known as 1-1, 2-1 and 2-2 electrolytes, respectively. The value of ‘A’ is the same for all electrolytes of the same type.
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संबंधित प्रश्न
Define “Molar conductivity”.
The conductivity of 0.20 M solution of KCl at 298 K is 0.025 S cm−1. Calculate its molar conductivity.
Define limiting molar conductivity.
10.0 grams of caustic soda when dissolved in 250 cm3 of water, the resultant gram molarity of solution is _______.
(A) 0.25 M
(B) 0.5 M
(C) 1.0 M
(D) 0.1 M
How can you determine limiting molar conductivity, 0 m for strong electrolyte and weak electrolyte?
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)
Kohlrausch law of independent migration of ions states ____________.
When acidulated water (dil.H2SO4 solution) is electrolysed, will the pH of the solution be affected? Justify your answer.
Assertion: `"E"_("Ag"^+ //"Ag")` increases with increase in concentration of Ag+ ions.
Reason: `"E"_("Ag"^+ //"Ag")` has a positive value.
Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The Λm of ‘B’ increases 1.5 times while that of A increases 25 times. Which of the two is a strong electrolyte? Justify your answer. Graphically show the behavior of ‘A’ and ‘B’.
An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to :-
Which of the following increases with the increase in the concentration of the solution?
The molar conductivity of CH3COOH at infinite dilution is 390 Scm2/mol. Using the graph and given information, the molar conductivity of CH3COOK will be:
Assertion (A) : Conductivity decreases with decrease in concentration of electrolyte.
Reason (R) : Number of ions per unit volume that carry the current in a solution decreases on dilution.
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.
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 unit of molar conductivity is ______.
Which of the following solutions will have the highest conductivity at 298 K?
The specific conductance of 2.5 × 10-4 M formic acid is 5.25 × 10-5 ohm-1 cm-1. Calculate its molar conductivity and degree of dissociation.
Given `λ°_("H"^+)` = 349.5 ohm-1 cm2 mol-1 and
`λ°_("HCOO"^-) = 50.5 " ohm"^-1 "cm"^2 "mol"^-1`
Suggest a way to determine the `∧_"m"^∘`value of water.
