20 ml of KOH solution was titrated with 0.2 M HCl solution in a conductivity cell. The data obtained was plotted to give the following graph. The concentration of the KOH solution was :

Engineering

Chemistry

Conductometric Titration

Question

20 ml of KOH solution was titrated with 0.2 M HCl solution in a conductivity cell. The data obtained was plotted to give the following graph.

The concentration of the KOH solution was :

0.1 M

0.2 M

0.4 M

0.8 M

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

At equivalent point (V_HCl used = 40 ml as shown in graph)
n_HCl = n_NaOH
$\frac{0 .2 \times 40}{1000} = \frac{M_{NaOH} \times 20}{1000}$
M_NaOH = 0.4

Please subscribe our Youtube channel to unlock this solution.