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Question
State Kohlrausch Law.
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Solution
Kohlrausch law states that at infinite dilution of the solution, each ion of electrolyte migrates independently of its co-ions and contribute independently to the total molar conductivity irrespective of the nature of other ion.
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RELATED QUESTIONS
Resistance of conductivity cell filled with 0.1 M KCl solution is 100 ohms. If the resistance of the same cell when filled with 0.02 M KCl solution is 520 ohms, calculate the conductivity and molar conductivity of 0.02 M KCl solution. [Given: Conductivity of 0.1 M KCl solution is 1.29 S m-1 .]
State Kohlrausch’s law of independent migration of ions.
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
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Why conductivity of an electrolyte solution decreases with the decrease in concentration ?
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.
The conductivity of 0.02 M AgNO3 at 25°C is 2.428 × 10−3 Ω−1 cm−1. What is its molar conductivity?
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.
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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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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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\[\ce{Λ^0_m H2O}\] is equal to:
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(ii) \[\ce{Λ^0_m_{(HNO_3)} + \ce{Λ^0_m_{(NaNO_3)} - \ce{Λ^0_m_{(NaOH)}}}}\]
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(iii) concentration of electrolytes in solution.
(iv) surface area of electrodes.
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`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 |
Assertion: `"E"_("Ag"^+ //"Ag")` increases with increase in concentration of Ag+ ions.
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Statement II: Molar conductivity decreases with decrease in concentration of electrolyte.
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`λ°_("HCOO"^-) = 50.5 " ohm"^-1 "cm"^2 "mol"^-1`
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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.
