10.0 ml of Na$$_2$$CO$$_3$$ solution is titrated against 0.2 M HCl solution. The following values were obtained in 5 readings. 4.8 ml, 4.9 ml, 5.0 ml, 5.0 ml and 5.0 ml
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)

In titrimetric analysis, we first identify and discard any outlier readings. The five readings are 4.8 mL, 4.9 mL, 5.0 mL, 5.0 mL, and 5.0 mL. By convention, the concordant readings (those that agree closely) are considered reliable. The three identical values of 5.0 mL are concordant, and the accepted titre value is taken as 5.0 mL.

Using titre = 5.0 mL of 0.2 M HCl for 10.0 mL of $$\text{Na}_2\text{CO}_3$$ solution, the reaction is $$\text{Na}_2\text{CO}_3 + 2\text{HCl} \to 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2$$.

Millimoles of HCl used $$= 0.2 \times 5.0 = 1.0$$ mmol. Since 2 moles of HCl react with 1 mole of $$\text{Na}_2\text{CO}_3$$, millimoles of $$\text{Na}_2\text{CO}_3 = \frac{1.0}{2} = 0.5$$ mmol. This 0.5 mmol is present in 10.0 mL of solution, so the concentration is $$\frac{0.5}{10.0} = 0.05$$ M $$= 50$$ mM.

The answer is $$50$$ mM.