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?

Options

A. 0.6 ampere
B. 0.214 ampere
C. 0.321 ampere
D. 0.1456 ampere

Answer

D

Explanation

`k = (^^_(eq)C)/(1000) = (97.1 xx 0.1)/(1000) = 9.71 xx 10^(-3) Scm^(-1)`
`G = (1)/(R) = k(A)/(l)`
`=9.71 xx 10^(-3) S cm^(-1) xx (1.5 cm^(2))/(0.5 cm) = 29.13 xx 10^(-3) ohm^(-1)`
`V = IR rArr I = (V)/(R) = 5 xx 29.13 xx 10^(-3)`
`=0.1456` amperes