The resistance of 0.25 M molar solution of an electrolyte was found to be 75 Ω.Calculate molar conductivity of the solution, if the electrodes in the cell are 1.8 cm apart and have an area of cross section 3.6 cm².

Strengthen your Class 12 Chemistry Electrochemistry concepts with this detailed numerical on molar conductivity. Learn how to calculate the molar conductivity of a 0.25 M electrolyte solution when its resistance is 75 Ω, and the conductivity cell has electrodes 1.8 cm apart with a 3.6 cm² cross-sectional area. This post explains the step-by-step procedure for determining the cell constant, conductivity, and final molar conductivity using standard formulas. Ideal for CBSE students and competitive exam aspirants looking for clear, accurate, and exam-focused electrochemistry solutions.

The resistance of 0.25 M molar solution of an electrolyte was found to be 75 Ω.Calculate molar conductivity of the solution, if the electrodes in the cell are 1.8 cm apart and have an area of cross section 3.6 cm².

Video solution:- 

The resistance of 0.25 M molar solution of an electrolyte was found to be 75 Ω.Calculate molar conductivit y of the solution, if the electrodes in the cell are 1.8 cm apart and have an area of cross section 3.6 cm².

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Questions:-

1.  Write the balanced half-reactions occurring at the anode and cathode of the galvanic cell:
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

2. Represent the cell in which the following reaction takes place

Mg(s)+2Ag+ (0.0001M)||Mg²+ (0.130M) + 2Ag(s)

Calculate its E (cell) if  E⁰ (cell) = 3.17V .

3.Calculate the emf of the cell in which the following reaction takes place:
Ni(s) + 2Ag+ (0.002 M) → Ni²+ (0.160 M) + 2Ag(s)
Given that E⁰cell = 1.05 V

4. A copper-silver cell is set up in which the concentration of copper ion is 0.10 M and the concentration of silver ion is not known. The cell potential when measured was found to be 0.422 V. Calculate the concentration of silver ions in the cell. The given cell reaction is:

Cu +2Ag+ → Cu²+ + 2Ag

(Given E°(Ag+/Ag) = +0.80V, E°(Cu²+/Cu) = +0.34 V), use antilog (1.29) = 19.5

5. Depict the galvanic cell in which the reaction
Zn(s) +2Ag+ (aq)  →  Zn²+ (aq) + 2Ag(s) takes place. Further show:
(i) Which of the electrode is negatively charged?
(ii) The carriers of the current in the cell.
(iii) Individual reaction at each electrode

6.Write the Nernst equation and emf of the following cells at 298 K:

Mg(s) |Mg²+ (0.001M)||Cu²+ (0.0001 M)|Cu(s)

7.Write the Nernst equation and emf of the following cells at 298 K:
(ii) Fe(s)|Fe²+ (0.001M)||H+(1M)| H2(g)(1bar)| Pt(s)

8.Write the Nernst equation and emf of the following cells at 298 K:

Sn(s)|Sn²+ (0.050M)||H+(0.020M)| H2(g)(1bar)| Pt(s)

9. Write the Nernst equation and emf of the following cells at 298 K:

Pt(s) | Br-(0.010M)|Br2(l)||H+(0.030M)| H2(g)(1bar)/ Pt(s).

10.Calculate the standard cell potentials of galvanic cell in which the following reactions take place:

                   2Cr(s)+3Cd²+ (aq)    →  2Cr³+ (aq) + 3Cd
Calculate the ΔG°  and equilibrium constant of the reactions.

11. Calculate the standard cell potentials of galvanic cell in which the following reactions take place:

                  Fe²+ (aq) + Ag+(aq)   →  Fe³+ (aq)+Ag(s)

Calculate the ΔG°  and equilibrium constant of the reactions.

12. What is the amount of electricity required to deposite one mole of aluminium from a solution   of AlCl3 ?

13.  A solution of Ni(NO3)2 is electrolysed between platinum electrode using a current of 5A for 30mint. What is the mass of Ni deposited at the cathode. The atomic mass of Ni=58.5g(amu). 

14.What is amount of electricity required to deposit one mole of calcium from molten  CaCl2 ?

15. What is amount of electricity required to deposit one mole of Zinc from ZnSO4?

16. Consider the reaction Cr2O7²- + 14H+  + 6e- → 2Cr2+   +  7H2O.  What is the quantity of electricity            in coulombs needed to reduce 1 mole of Cr2O72-  ? 

17.Calculate the mass of silver deposited from silver nitrate solution by a current of 2 ampere                 flowing for 30 minutes. (Equivalent mass of silver is 108).

18. How many hours does it take to reduce 3 mol of Fe3+ to Fe2+ with 2.00 A current? (1 Faraday = 96,500 C mol-¹, R = 8.314JK-1 mol-1 ) 

19. What mass of zinc can be produced by electrolysis of zinc sulphate solution when current of 1.5 Ampere is passed for 15 minutes ?The atomic Mass of Zinc = 65.4 amu. 

20. If a current of 0.5 ampere flows through a metallic wire for 2 hours, then howmany electrons would flow through the wire?

21. The molar conductance of infinite dilution for sodium acetate ( CH3COONa) Hydrochloric acid (HCI) and sodium chloride (NaCl) are 92.5, 426.9 and 120.4 S cm²mol-¹ respectively at 298 K. calculate the molar conductance of Acetic Acid (CH3COOH) at infinite dilution.

22.Calculate the molar conductance of a solution of MgCl2 at infinite dilution, given that molar ionic Conductance of ∧⁰ (Mg²+) = 126.1Scm²mol-¹ and ∧⁰(Cl) = 56.3 Scm²mol-¹

23. Calculte molar conductance at infinite dilution for acetic acid, given ∧⁰HCI=425ohm−¹cm²mol−¹,∧⁰ NaCI=188ohm−¹cm²mol−¹,∧⁰  CH3COONa=96ohm−¹cm²mol−¹ 

24.  Given molar conductivities at infinite dilution ∧⁰m for Ba(OH)2 = 517.6 Ω−¹cm²mol−¹ ∧⁰m for BaCl2 = 240.6 Ω−¹cm²mol−¹, ∧⁰m for NH4Cl =  129.8Ω−¹cm²mol−¹ . Calculate ∧⁰m for NH4OH

25. The resistance of 0.25 M molar solution of an electrolyte was found to be 75 Ω.Calculate molar conductivity of the solution, if the electrodes in the cell are 1.8 cm apart and have an area of cross section 3.6 cm². 

26. The Resistance of 0.5 M solution of an electrolyte was found to be 30 Ω. Calculate the molar conductivity of the solution if the electrodes in the cell are 1.5 cm apart and having area of cross section 2.0 cm². 

27. The electrical Resistance of a column of 0.05 M NaOH solution of diameter 1cm and length 50 cm is 5.55 × 10³ Ω. Calculate its Molar conductivity.

28. The Resistance of a 0.5 N solution of an electrolyte in a conductivity cell was found to be 25 Ω. Calculate the equivalent conductivity if the electrodes in the cell are 1.6 cm apart and having an area cross section 3.2 cm². 

29. Write equation and calculate e.m.f of cell at 298 K.

Mg(s) / Mg²+ (0.001M)|| Cu²+ (0.0001M)/Cu(s)

Given E⁰Mg²+/Mg(s) = - 2.37 V, E⁰cu²+/Cu(s) = 0.34 V

30. Calculate ΔG, ΔG° and Ecell for cell.

Al/Al³+ (0.01M)||Fe²+ (0.02M) / Fe
Given that E⁰ (Fe²+/ Fe ) = -0.44V, E⁰ (Al³+/Al ) =-1.66V
F = 96500 C,  R = 8.314JK-¹ mol-¹

31. Calculate Ecell and ΔG for cell,

Cu (s)/ Cu²+ (0.130M) || Ag+ / Ag(s) (1.0×10-⁴M)
E⁰Cu+/Cu = 0.34 V, E⁰Ag+/Ag = 0.80 V
Given, F = 96500 C, R = 8.314 JK-1 mol-1 

32. Write the Nernst equation and calculate the e.m.f. of the following cell at 298 K:

Mg (s) |Mg²+ (0.001M)||Cu²+ (0.0001M)|Cu (s)
Given E⁰ Mg²+ / Mg =-2.37V, E⁰ Cu²+ /Cu =0.34 V 

33. Write the Nernst equation and calculate e.m.f of the following cell of 298 K.

Ni|Ni²+ (0.01M)||Cu²+ | Cu(0.01M)
Given E⁰(Cu²+/Cu) = 0.34V, E⁰ (Ni²+ /Ni) =-0.22V 

34. Write the nernst equation and Calculate the e.m.f of the cell of 298 K.

Mg(s)|Mg²+ (0.130M)| |Ag+(1.0×10-⁴M)|Ag(s)
Given, E⁰ Ag+ /Ag =0.80 V, E (Mg²+ / Mg ) =-2.36 V 

35. Calculate the e.m.f. of the following cell.

Cr | Cr³+ (0.1M) || Fe²+ (0.01M) | Fe
Given,  E⁰(Cr³+/Cr) = -0.75V,   E⁰(Fe²+ /Fe) = -0.45V