Question

The specific conductance of saturated solution of $Ca{F_2}$ is $3.80 \times {10^{ - 5}}\,mho\,c{m^{ - 1}}$ and that water used for is$0.15 \times {10^{ - 5}}$. The specific conductance of $Ca{F_2}$ alone is,
A.$3.71 \times {10^{ - 5}}$
B.$4.01 \times {10^{ - 5}}$
C.$3.7 \times {10^{ - 4}}$
D.$3.86 \times {10^{ - 4}}$

Answer 1

Hint: We can define specific conductance as the ability of the material to conduct electricity. The specific conductance is expressed as \[k\]. The reciprocal of resistance is conductance. We can calculate the specific conductance of $Ca{F_2}$ by subtracting the specific conductance of saturated solution to the specific conductance of water.
Formula used: We can calculate the specific conductance of $Ca{F_2}$ as,
$k\left( {Ca{F_2}} \right) = k\left( {saturated\,solution} \right) - k\left( {{H_2}O} \right)$
Here, $k\left( {Ca{F_2}} \right)$ is the specific conductance of $Ca{F_2}$.
$k\left( {saturated\,solution} \right)$ is the specific conductance of the saturated solution.
$k\left( {{H_2}O} \right)$ is the specific conductance of water.

Complete step by step answer:
We can see the given data contains specific conductance of saturated solution is $3.80 \times {10^{ - 5}}\,{\Omega ^{ - 1}}\,c{m^{ - 1}}$ and specific conductance of water is $0.15 \times {10^{ - 5}}\,{\Omega ^{ - 1}}\,c{m^{ - 1}}$.
We know that the specific conductance of saturated solution is the sum of specific conductance of $Ca{F_2}$ and sum of specific conductance of water. We can write the equation as,
\[k\left( {saturated\,solution} \right) = k\left( {Ca{F_2}} \right) + k\left( {{H_2}O} \right)\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\]
Here, $k\left( {Ca{F_2}} \right)$ is the specific conductance of $Ca{F_2}$.
$k\left( {saturated\,solution} \right)$ is the specific conductance of the saturated solution.
$k\left( {{H_2}O} \right)$ is the specific conductance of water.
We can rearrange the equation (1) to get the specific conductance of $Ca{F_2}$. We write the rearranged equation as,
$k\left( {Ca{F_2}} \right) = k\left( {saturated\,solution} \right) - k\left( {{H_2}O} \right)\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)$
We can substitute the values of specific conductance of saturated solution and specific conductance of water in equation (2) to calculate the specific conductance of $Ca{F_2}$.
The value of $k\left( {saturated\,solution} \right)$ is $3.80 \times {10^{ - 5}}\,{\Omega ^{ - 1}}\,c{m^{ - 1}}$.
The value of $k\left( {water} \right)$ is $0.15 \times {10^{ - 5}}\,{\Omega ^{ - 1}}\,c{m^{ - 1}}$.
Now we can calculate the specific conductance of $Ca{F_2}$ as,
$
  k\left( {Ca{F_2}} \right) = k\left( {saturated\,solution} \right) - k\left( {{H_2}O} \right) \\
  k\left( {Ca{F_2}} \right) = k\left( {3.86 - 0.15} \right) \times {10^{ - 5}}\,{\Omega ^{ - 1}}\,c{m^{ - 1}} \\
  k\left( {Ca{F_2}} \right) = 3.71 \times {0^{ - 5}}\,{\Omega ^{ - 1}}\,c{m^{ - 1}} \\
 $
The calculated specific conductance of $Ca{F_2}$ is $3.71 \times {10^{ - 5}}{\Omega ^{ - 1}}c{m^{ - 1}}$.
$\therefore $ Option (A) is correct.

Note:
In simple words, the reciprocal of resistivity is called specific conductivity. We could define specific conductivity as the conductance between the opposite faces of the centimeter cube of a conductor. We can represent it as $k$ (kappa). We can write the expression of specific conductivity as,
$k = \dfrac{1}{\rho }$
Here, $\rho $ is the resistivity of the material.
The SI unit of specific conductance is Siemens. Siemens is the reciprocal of one ohm and it is represented as mho. We can write the symbol of Siemens as $S\,\left( {or} \right)\,{\Omega ^{ - 1}}$.