Question

A $\mathrm{KCl}$ solution of conductivity $0.14\mathrm{S}{\mathrm{m}}^{-1}$shows a resistance of $4.19\mathrm{ohm}$ in a conductivity cell. If the same cell is filled with an $\mathrm{HCl}$ solution, the resistance drops $1.03\Omega$. The conductivity of the $\mathrm{HCl}$ solution is $.............\times {10}^{-2}\mathrm{S}{\mathrm{m}}^{-1}$. (Round off to the Nearest Integer).

Answer 1

1. The conductivity of an electrolytic solution is its ability to conduct electricity.
2. Conductivity is reciprocal of resistivity.
3. It is given that we have two electrolytic solutions $\mathrm{KCl}$ and $\mathrm{HCl}$.
4. The conductivity of $\mathrm{KCl}$is $0.14\mathrm{S}{\mathrm{m}}^{-1}$, resistance of $\mathrm{KCl}$ is $4.19\Omega$ and resistance of $\mathrm{HCl}$ is $1.03\Omega$.
5. The conductivity of an electrolytic solution is determined by the formula: $\mathrm{R}=\frac{\mathrm{l}}{\mathrm{\kappa }\mathrm{A}}$ where, $\mathrm{R}$ is resistance, $\kappa$is conductivity, $\mathrm{l}$ is the length of material containing the solution and $\mathrm{A}$ is the area of cross-section of the container.
6. Putting the values for $\mathrm{KCl}$ solution in the value. We get: $$\begin{gathered} 4.19=\frac{\mathrm{l}}{0.14\mathrm{A}} \\ 4.19\times 0.14=\frac{\mathrm{l}}{\mathrm{A}} \\ \frac{\mathrm{l}}{\mathrm{A}}=0.58\Omega \end{gathered}$$
7. Now, placing the values of $\mathrm{R}$ and $\frac{\mathrm{l}}{\mathrm{A}}$in the formula to determine the conductivity of $\mathrm{HCl}$ solution: $$\begin{gathered} 1.03=\frac{0.58}{\kappa } \\ \kappa =0.56\mathrm{S}{\mathrm{m}}^{-1} \\ =56\times {10}^{-2}\mathrm{S}{\mathrm{m}}^{-1} \end{gathered}$$
8. So, the conductivity of $\mathrm{HCl}$ solution is $\mathbf{56}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}\mathbf{}}\mathbf{S}\mathbf{}{\mathbf{m}}^{\mathbf{-}\mathbf{1}}$.