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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.
संबंधित प्रश्न
The conductivity of 0.20 M solution of KCl at 298 K is 0.025 S cm−1. Calculate its molar conductivity.
The conductivity of 0.001 mol L-1 solution of CH3COOH is 3.905× 10-5 S cm-1. Calculate its molar conductivity and degree of dissociation (α) Given λ°(H+)= 349.6 S cm2 mol-1 and λ°(CH3COO)= 40.9S cm2mol-1.
Define limiting molar conductivity.
Define the following terms: Molar conductivity (⋀m)
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
Define the following terms :
Limiting molar conductivity
How can you determine limiting molar conductivity, 0 m for strong electrolyte and weak electrolyte?
The S.I. unit of cell constant for conductivity cell is __________.
\[\ce{\Lambda^0_m(NH4OH)}\] is equal to ______.
When acidulated water (dil.H2SO4 solution) is electrolysed, will the pH of the solution be affected? Justify your answer.
Match the items of Column I and Column II on the basis of data given below:
`E_("F"_2//"F"^-)^Θ` = 2.87 V, `"E"_(("Li"^(+))//("Li"^-))^Θ` = − 3.5V, `"E"_(("Au"^(3+))//("Au"))^Θ` = 1.4 V, `"E"_(("Br"_(2))//("Br"^-))^Θ` = 1.09 V
| Column I | Column II |
| (i) F2 | (a) metal is the strongest reducing agent |
| (ii) Li | (b) metal ion which is the weakest oxidising agent |
| (iii) Au3+ | (c) non metal which is the best oxidising agent |
| (iv) Br– | (d) unreactive metal |
| (v) Au | (e) anion that can be oxidised by Au3+ |
| (vi) Li+ | (f) anion which is the weakest reducing agent |
| (vii) F– | (g) metal ion which is an oxidising agent |
Which of the following halogen acids is the strongest reducing agent?
Given below are two statements:
Statements I: The limiting molar conductivity of KCl (strong electrolyte) is higher compared to that of CH3COOH (weak electrolyte).
Statement II: Molar conductivity decreases with decrease in concentration of electrolyte.
In the light of the above statements, choose the most appropriate answer from the options given below:
The solubility of Co2[Fe(CN)6] in water at 25°C from the following data:
Conductivity of saturated solution of Co2[Fe(CN)6] = 2.06 × 10−6 ohm−1 cm−1 and that of water = 4.1 × 10−7 ohm−1 cm−1. The ionic molar conductivities of Co2+ and [Fe(CN)6]4− are 86 and 444 ohm−1 cm2 mol−1 respectively, is ______ × 10−6 mol/L.
The variation of molar conductivity with concentration of an electrolyte (X) m aqueous solution is shown in the given figure.
The electrolyte X is ______.
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 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`
The resistance of a conductivity cell with a 0.1 M KCl solution is 200 ohm. When the same cell is filled with a 0.02 M NaCl solution, the resistance is 1100 ohm. If the conductivity of 0.1 M KCl solution is 0.0129 ohm-1 cm-1, calculate the cell constant and molar conductivity of 0.02 M NaCl solution.
