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प्रश्न
Suggest a way to determine the \[\ce{\Lambda^{\circ}_m}\] value of water.
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उत्तर
Water is a weak electrolyte. Its \[\ce{\Lambda^{\circ}_m}\] value can be determined with the help of Kohlrausch’s law.
\[\ce{\Lambda^{\circ}_m(HCl) = \Lambda^{\circ}_m(H+) + \Lambda^{\circ}_m(Cl^-)}\] ...(i)
\[\ce{\Lambda^{\circ}_m(NaOH) = \Lambda^{\circ}_m(Na+) + \Lambda^{\circ}_m(OH^-)}\] ...(ii)
\[\ce{\Lambda^{\circ}_m(NaCl) = \Lambda^{\circ}_m(Na+) + \Lambda^{\circ}_m(Cl^-)}\] ...(iii)
Adding (i) and (ii) and subtracting (iii), we get,
\[\ce{\Lambda^{\circ}_m(H2O) = \Lambda^{\circ}_m(H+) + \Lambda^{\circ}_m(OH^-)}\]
= \[\ce{\Lambda^{\circ}_m(HCl) + \Lambda^{\circ}_m(NaOH) - \Lambda^{\circ}_m(NaCl)}\]
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संबंधित प्रश्न
The molar conductivity of cation and anion of salt BA are 180 and 220 mhos respectively. The molar conductivity of salt BA at infinite dilution is_____________ .
(a) 90 mhos.cm2
(b) 110 mhos.cm2.mol-1
(c) 200 mhos.cm2.mol-1
(d) 400 mhos.cm2.mol-1
State Kohlrausch Law.
Define limiting molar conductivity.
The conductivity of 0.20 mol L−1 solution of KCl is 2.48 × 10−2 S cm−1. Calculate its molar conductivity and degree of dissociation (α). Given λ0 (K+) = 73.5 S cm2 mol−1 and λ0 (C1−) = 76.5 S cm2 mol−1.
Why does the conductivity of a solution decrease with dilution?
The molar conductivity of 0.025 mol L−1 methanoic acid is 46.1 S cm2 mol−1. Calculate its degree of dissociation and dissociation constant. Given \[\ce{λ^0_{(H^+)}}\] = 349.6 S cm2 mol−1 and \[\ce{λ^0_{(HCOO^-)}}\] = 54.6 S cm2 mol−1.
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)
How can you determine limiting molar conductivity, 0 m for strong electrolyte and weak electrolyte?
In the plot of molar conductivity (∧m) vs square root of concentration (c1/2) following curves are obtained for two electrolytes A and B : 
Answer the following:
(i) predict the nature of electrolytes A and B.
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(ii) distance between electrodes.
(iii) concentration of electrolytes in solution.
(iv) surface area of electrodes.
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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
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