Answer
$c(HCl) = 0.2331M$
Work Step by Step
Considering stoichiometric coefficients in the reaction given, the amount of hydrochloric acid required is twice the amount of sodium - carbonate. Hence,
$n(HCl) = 2n(Na_{2}CO_{3}) = 2\times \frac{m(Na_{2}CO_{3})}{M(Na_{2}CO_{3})} = 2\times \frac{0.483g}{106\frac{g}{mol}}= 0.0091132mol$
Given the volume of $HCl$ solution, we can easily obtain its concentration:
$c(HCl) = \frac{n(HCl)}{V(HCl)} = \frac{0.0091132mol}{0.0391dm^3}=0.2331M$
