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Question: The specific conductance of saturated solution of \(Ca{F_2}\) is \(3.80 \times {10^{ - 5}}\,mho\,c{m...

The specific conductance of saturated solution of CaF2Ca{F_2} is 3.80×10−5 mho cm−13.80 \times {10^{ - 5}}\,mho\,c{m^{ - 1}} and that water used for is0.15×10−50.15 \times {10^{ - 5}}. The specific conductance of CaF2Ca{F_2} alone is,
A.3.71×10−53.71 \times {10^{ - 5}}
B.4.01×10−54.01 \times {10^{ - 5}}
C.3.7×10−43.7 \times {10^{ - 4}}
D.3.86×10−43.86 \times {10^{ - 4}}

Explanation

Solution

We can define specific conductance as the ability of the material to conduct electricity. The specific conductance is expressed as kk. The reciprocal of resistance is conductance. We can calculate the specific conductance of CaF2Ca{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 CaF2Ca{F_2} as,
k(CaF2)=k(saturated solution)−k(H2O)k\left( {Ca{F_2}} \right) = k\left( {saturated\,solution} \right) - k\left( {{H_2}O} \right)
Here, k(CaF2)k\left( {Ca{F_2}} \right) is the specific conductance of CaF2Ca{F_2}.
k(saturated solution)k\left( {saturated\,solution} \right) is the specific conductance of the saturated solution.
k(H2O)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 3.80×10−5 Ω−1 cm−13.80 \times {10^{ - 5}}\,{\Omega ^{ - 1}}\,c{m^{ - 1}} and specific conductance of water is 0.15×10−5 Ω−1 cm−10.15 \times {10^{ - 5}}\,{\Omega ^{ - 1}}\,c{m^{ - 1}}.
We know that the specific conductance of saturated solution is the sum of specific conductance of CaF2Ca{F_2} and sum of specific conductance of water. We can write the equation as,
k(saturated solution)=k(CaF2)+k(H2O)          (1)k\left( {saturated\,solution} \right) = k\left( {Ca{F_2}} \right) + k\left( {{H_2}O} \right)\,\,\,\,\,\,\,\,\,\,\left( 1 \right)
Here, k(CaF2)k\left( {Ca{F_2}} \right) is the specific conductance of CaF2Ca{F_2}.
k(saturated solution)k\left( {saturated\,solution} \right) is the specific conductance of the saturated solution.
k(H2O)k\left( {{H_2}O} \right) is the specific conductance of water.
We can rearrange the equation (1) to get the specific conductance of CaF2Ca{F_2}. We write the rearranged equation as,
k(CaF2)=k(saturated solution)−k(H2O)            (2)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 CaF2Ca{F_2}.
The value of k(saturated solution)k\left( {saturated\,solution} \right) is 3.80×10−5 Ω−1 cm−13.80 \times {10^{ - 5}}\,{\Omega ^{ - 1}}\,c{m^{ - 1}}.
The value of k(water)k\left( {water} \right) is 0.15×10−5 Ω−1 cm−10.15 \times {10^{ - 5}}\,{\Omega ^{ - 1}}\,c{m^{ - 1}}.
Now we can calculate the specific conductance of CaF2Ca{F_2} as,
k(CaF2)=k(saturated solution)−k(H2O) k(CaF2)=k(3.86−0.15)×10−5 Ω−1 cm−1 k(CaF2)=3.71×0−5 Ω−1 cm−1  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 CaF2Ca{F_2} is 3.71×10−5Ω−1cm−13.71 \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 kk (kappa). We can write the expression of specific conductivity as,
k=1ρk = \dfrac{1}{\rho }
Here, ρ\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 (or) Ω−1S\,\left( {or} \right)\,{\Omega ^{ - 1}}.