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The equivalent conductance of 0.10 N sol...

The equivalent conductance of `0.10 N` solution of `MgCI_(2)` is 97.1 mho `cm^(2) eq^(-1)`. A cell electrodes that are `1.50 cm^(2)` in surface are and `0.50` cm apart is filled with `0.1N MgCI_(2)` solution. How much current will flow when the potential difference between the electrodes is 5 volts?

Text Solution

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Cell constant `= (0.50)/(1.50) = (1)/(3)`
Specific conductance `= ("equivalent conductance")/("volume (c c) containing 1 eq.")`
`= (97.1)/(10,000)` (for 0.1 N solution V = 10,000 cc)
`= 0.00971 "mho cm^(-1)`.
Conductance = specific conductance/cell constant
`= (0.00971)/(1//3) = 0.02913` mho.
`therefore` resistance `= (1)/(0.02913)` ohm.
`therefore` current in amp `= ("potential difference (volt)")/("resistance (ohm)")` (Ohm.s law)
`= (5)/(1//0.02913) = 0.1456` ampere.
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