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Question

The equivalent conductance of 0.1 N solution of MgCl2 is 25 Ω−1cm2eq−1 at 25 ∘C. A cell with electrodes that are 2 cm2 in surface area and 0.5 cm apart is filled with 0.1 N MgCl2 solution. How much current (in A) will flow when the potential difference between the electrodes is 5 Volts?

A
0.05
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B
0.01
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C
0.5
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D
0.1
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Solution

The correct option is A 0.05Cell constant (la)=0.52=14 Specific conductance (K) =Equivalent conductance Volume (cc) containing 1 eq.=2510000 (∵For 0.1 N solution volume =10000 cc)=0.0025 Ω−1cm−1 Conductance =K×al=0.0025×4=0.01 Ω−1 ∴ Resistance (R)=10.01=100 Ω ∴ Current (in A) =Potential diffrence (V)Resistance (Ω)=5100=0.05 A

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