10 . 0ml of Na _ 2 CO _ 3 solution is titrated against 0 . 2MHCl...

$10 . 0ml$ of ${Na}_{2}{CO}_{3}$ solution is titrated against $0 . 2MHCl$ solution. The following values were obtained in $5$ readings. $4 . 8ml, 4 . 9ml, 5 . 0ml, 5 . 0ml$ and $5 . 0ml$

Based on these readings, and convention of titrimetric estimation of concentration of ${Na}_{2}{CO}_{3}$ solution is ___ $mM$ .
(Round off to the Nearest integer)

Correct answer

Step-by-step solution

Most precise volume of $HCl = 5ml$

at equivalence point

Meq. of ${Na}_{2}{CO}_{3} =$ meq. of $HCl$

Let molarity of ${Na}_{2}{CO}_{3}$

solution $= M$ , then

$M \times 10 \times 2 = 0 . 2 \times 5 \times 1$

$M = 0 . 05mol / L$

$= 0 . 05 \times 1000$

$= 50mM$

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