Question

The equivalent conductance of 0.10 N solution of $\text{MgC}{\text{l}}_{\text{2}}$ is 97.1 mho $\text{c}{\text{m}}^2\text{equ}{\text{i}}^{-1}$ at ${25}^{\circ }\text{C}$. A cell with electrode that are 1.5 $\text{c}{\text{m}}^2$ in surface area and 0.5 cm apart is filled with 0.1 N $\text{MgC}{\text{l}}_2$ solution. How much amperes of current will flow when potential difference between the electrodes is 5 volt?

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

Answer 1

The correct option is D 0.1456

Given ${\Delta }_{eq}=97.1\text{mho}\text{c}{\text{m}}^2\text{equ}{\text{i}}^{-1}$
Concentration of $\text{MgC}{\text{l}}_{\text{2}}=0.10\text{N}=\text{C}$
Surface Area (A) =1.5 $\text{c}{\text{m}}^2$
length (l) = 0.5 cm
Now, ${\Delta }_{eq}=\text{K}\times \frac{1000}{\text{C}}$
$97.1=\text{K}\times \frac{1000}{0.1}$
K = 0.0097 mho $\text{c}{\text{m}}^2$
Now, $\text{R}=\rho \frac{l}{\text{A}}$
Since $\rho =\frac{1}{\text{K}}=\frac{1}{0.0097}=103.093\text{ohm}\text{.}\text{cm}$
$\text{R}=\text{103}\text{.093}\times \frac{0.5}{1.5}=34.36\text{ohm}$
According to ohm's Law
V = IR
V = 5 volt, R = 3.36 ohm
$5=\text{I}\times 34.36$
$\text{I}=0.145\text{Ampere}$
Hence option (D) is correct.