Question

A conductivity cell filled with 0.02 M ${{H}_{2}}S{{O}_{4}}$ gives at ${{25}^{\circ }}C$ a resistance of 122 ohms. If the molar conductivity of 0.02 M ${{H}_{2}}S{{O}_{4}}$ is $618{{\Omega }^{-1}}$, what is the cell constant?
A. $1.2c{{m}^{-1}}$
B. $1.3c{{m}^{-1}}$
C. $1.4c{{m}^{-1}}$
D. $1.5c{{m}^{-1}}$

Answer 1

Conductivity is basically an ability of any substance to allow current to pass through it. As we know that the cell constant has a unit of per centimetre and is denoted by the symbol ‘k’. We can determine the cell constant if any solution with the known conductivity is given.

Complete Step by step solution:
- As we know that the cell constant is the ratio of the distance in between the two electrodes and the area of the electrodes.
- We are being provided with the values of:
Resistance (R)= 122 ohms
Molar conductivity (${{\lambda }_{m}}$) = $618{{\Omega }^{-1}}$
Concentration (C)= 0.02 M
- We have to find the value of Cell constant (x)
As we know that the molar conductivity is given by the equation:
${{\lambda }_{m}}=\dfrac{1000\times k}{C}$
Where, k is the specific conductivity.
Now, by putting all the values in the above equation we get:

& 618=\dfrac{1000\times k}{0.02} \\\ & k=\dfrac{618\times 0.02}{1000} \\\ & k=0.01236\text{ }{{\Omega }^{-1}}c{{m}^{-1}} \\\ \end{aligned}$$ Where, k is the conductivity. $$k=Gx$$ $$G=\dfrac{1}{R}$$ (x) is Cell constant $$\begin{aligned} & 0.01236=\dfrac{1}{R}x \\\ & 0.01236=\dfrac{1}{122}x \\\ & 1.50792=x \\\ & x=1.5\text{ }c{{m}^{-1}} \\\ \end{aligned}$$ \- The cell constant is found to be $$x=1.5\text{ }c{{m}^{-1}}$$ **\- Hence, we can conclude that the correct option is (d), that is the cell constant is $1.5c{{m}^{-1}}$.** **Note:** \- It is found that when the conductivity of an electrode is high, then the value of cell constant (k) is less than one. Whereas, if the conductivity of an electrode is low then the value of conductivity can increase up to 1.