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# The equivalent conductance of 0.10 N solution of $\text{MgC}{{\text{l}}_{\text{2}}}$ is $\text{97}\text{.1mho c}{{\text{m}}^{\text{2}}}\text{ equ}{{\text{i}}^{\text{-1}}}$ at $\text{2}{{\text{5}}^{\text{o}}}\text{C}$. A cell with electrode that are $\text{1}\text{.5c}{{\text{m}}^{\text{2}}}$ in surface area and 0.5cm apart if filled with 0.1 N $\text{MgC}{{\text{l}}_{\text{2}}}$ solution. How much amperes of current will flow when the potential difference between the electrode is 5 volts?(a) 0.6(b) 0.214(c) 0.321(d) 0.1456

Last updated date: 13th Jun 2024
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Hint: First the cell constant is calculated by dividing length and area of the conductor. Then specific conductance in which the volume should be taken in a cubic centimeter. Conductance is calculated with both. Ohm's is applied for the solution. Current can be calculated by dividing potential difference with resistance.

The area of the electrode is given = $\text{1}\text{.5c}{{\text{m}}^{\text{2}}}$
The length is also given = 0.5 cm
So, with these we can calculate the cell constant,
$\text{cell constant=}\dfrac{\text{l}}{\text{a}}$= $\dfrac{\text{0}\text{.50}}{\text{1}\text{.50}}\text{=}\dfrac{\text{1}}{\text{3}}$
Now, let us calculate the specific conductance with equivalent conductance and volume in cc (cubic centimeter)
$\text{specific conductance=}\dfrac{\text{equivalent conductance}}{\text{volume(cc)}}$
Given the equivalent conductance = $\text{97}\text{.1mho c}{{\text{m}}^{\text{2}}}\text{ equ}{{\text{i}}^{\text{-1}}}$
Volume for the 0.1 N solution in cc will be = 10000 cc
$\text{specific conductance=}\dfrac{\text{97}\text{.1}}{\text{10000}}$$\text{=0}\text{.00971mho c}{{\text{m}}^{\text{-1}}}$
Now, we have the specific conductance and the cell constant, we can calculate the conductance.
Conductance is given by:
$\text{conductance=}\dfrac{\text{specific conductance}}{\text{cell constant}}$
Putting the values, $\text{conductance=}\dfrac{\text{0}\text{.00971}}{\text{1/3}}\text{=0}\text{.02931}$
Now, resistance is reciprocal of conductance.
The resistance will be
$\text{resistance=}\dfrac{\text{1}}{\text{conductance}}\text{=}\dfrac{\text{1}}{\text{0}\text{.02931}}$
Now, for calculating the current in ampere “ohm’s law” should be used.
According to ohm’s law, the current can be calculated by dividing the potential difference with resistance,
$\text{current=}\dfrac{\text{potential difference}}{\text{resistance}}$
$\text{current=}\dfrac{\text{5}}{\text{1/0}\text{.02931}}\text{=0}\text{.1456}$
Hence, the current in ampere that will flow is 0.1456 amperes.
So, the correct option is (d) 0.1456.

Note: While calculating any quantity the dimension should be checked and they should be converted accordingly. Here the volume of the solution is taken in cc not in liters because the specific conductance is in cm cube.