Question

Conductivity (unit Siemens’ S) is directly proportional to area of the vessel and the concentration of the solution in it and is inversely proportional to the length of the vessel then the unit of the constant of proportionality is _________.
a.) \[(S)(m){(mol)^{ - 1}}\]
b.) \[\left( S \right){\left( m \right)^2}{\left( {mol} \right)^{ - 1}}\]
c.) \[{\left( S \right)^{ - 2}}{\left( m \right)^2}{\left( {mol} \right)^{ - 1}}\]
d.) \[{\left( S \right)^2}{\left( m \right)^2}{\left( {mol} \right)^{ - 1}}\]

Answer 1

Hint: In order to deal with this question first we will know about the term conductivity further we will make a formula according to the question statement and we will write the SI units of all the terms to evaluate the required answer.
Conductivity: Conductivity is nothing more than the indicator of the water's capacity to carry electric current surge. This conductance potential is said to be directly proportional to the ion concentration present in the water.

Complete step by step answer:
Now according to the question statement:
Let S denote conductivity.
As we have: conductivity is directly proportional to area of the vessel
$ \Rightarrow S \propto A$---------equation (1), where $A = $area

As we have: conductivity is directly proportional to the concentration of the solution in the vessel
$ \Rightarrow S \propto C$---------equation (2), where $C = $concentration

As we have: conductivity is inversely proportional to the length of the vessel
$ \Rightarrow S \propto \dfrac{1}{L}$---------equation (3), where $L = $length

By combining all the proportionality from equation (1), (2) and (3) we have:
$ \Rightarrow S \propto \dfrac{{AC}}{L}$
So, the formula becomes:
$ \Rightarrow S = K\dfrac{{AC}}{L}$, where $K = $constant of proportionality.
$ \Rightarrow K = \dfrac{{SL}}{{AC}}$
As we know that the SI units of following terms are:
SI unit of conductivity:$ = S$, $[S = siemens's]$
SI unit of length: $m$, $[m = meter]$
SI unit of area: ${m^2}$, $[m = meter]$
SI unit of concentration: $\dfrac{{mol}}{{{m^3}}}$, $[mol = mole]$

Now substitute the SI units of all the terms to find the unit of K, so we get:
Since, $K = \dfrac{{SL}}{{AC}}$
${(unit)_K} = \dfrac{{S \times m}}{{{m^2} \times \dfrac{{mol}}{{{m^3}}}}}$
$ \Rightarrow {(unit)_K} = \dfrac{{S \times m \times {m^3}}}{{{m^2} \times mol}}$
$ \Rightarrow {(unit)_K} = S \times {m^2} \times mo{l^{ - 1}}$
$ \Rightarrow {(unit)_K} = S{\text{ }}{m^2}{\text{ }}mo{l^{ - 1}}$
Hence, the unit of the constant of proportionality is \[\left( S \right){\left( m \right)^2}{\left( {mol} \right)^{ - 1}}\].
So, the correct answer is “Option B”.

Note: Here we have seen that the given compound is having some concentration that means it can be dissolved, and the conductivity of an electrolyte is dependent on the area of the vessel and the concentration of the solution and the length of the vessel in which it is placed. Hence the medium which is giving free electrons and behaves like a conductive medium are the electrolytes. So according to the amount of the ions there are some strong electrolyte (which is having large number of ions, e.g table salt and sodium chloride) and some are weak electrolyte (which is having less number of ions, e.g ammonia, acetic acid). Conductivity sometimes also known as specific conductance. If some of the salts are dissolved in an electrolyte then the conductance will increase drastically.