Question

The eqivalent conductance of 0.10 N solution of $MgC{l}_2$ is $97.1mhoc{m}^{2}e{q}^{-1}$. A cell with electrodes that are $1.50c{m}^2$ in surface area and 0.50 cm apart is filled with $0.1N$ $MgC{l}_2$ solution. The current (in amp) that will flow when the potential difference between the electrodes is 5 volts is $145\times {10}^{-x}$, then what is the value of x?

Answer 1

Cell constant $=\frac{0.50}{1.50}=\frac{1}{3}$

Specific conductance $=\frac{equivalentconductance}{\text{volume (cc) containing 1 eq}.}$
$=\frac{97.1}{10,000}$ (for 0.1 N solution $V=10,000cc$)
$=0.00971mhoc{m}^{-1}$

Conductance $=$ specific conductance/ cell constant

$\frac{0.00971}{1/3}=0.02913mho$

$\therefore resistance=\frac{1}{0.02913}ohm$

$\therefore$ current in amp $=\frac{potentialdifference\left(volt\right)}{resistance\left(ohm\right)}$

(Ohm's law) $=\frac{5}{1/0.02913}=0.1456ampere=145.6\times {10}^{-3}A$.