`20 Ml` of `KOH` solution was titrated with `0.20 M H_(2)SO_(4)` solution in a conductivity cell. The data obtained were plotted to give the graphe shown below.
The concentration of the `KOH` solution was

Options

A. `0.3M`
B. `0.15M`
C. `0.12M`
D. `0.075M`

Answer

A

Explanation

`20 xx M = 0.2 xx 2xx 15 rArr M = 0.15`