25 mL of HCl solution is titrated with 0.10 mole L–1 NaOH solution in a conductivity cell. The data obtained were plotted to give the graph shown below The concentration of HCl solution is

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25 mL of HCl solution is titrated with 0.10 mole L^–1 NaOH solution in a conductivity cell. The data obtained were plotted to give the graph shown below

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The concentration of HCl solution is

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0.060 M

0.040 M

0.080 M

0.01 M

answer is C.

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Detailed Solution

When 20 ml of NaOH is added, equivalents point is reached.

Equivalents of HCl = Equivalents of NaOH

25 x 10^-3 x M = 10^-1 x 20 x 10^-3

$M = \frac{20}{25 \times 10} = 0.08 M$

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