Answer
$c(HNO_{3})=0.1969 \frac{mol}{dm^{3}}$
Work Step by Step
$V(HNO_{3})=19.55ml=0.01955dm^{3}$
$m(Na_{2}CO_{3})=0.2040g$---$c(HNO_{3})=?$
The neutralization reaction of $Na_{2}CO_{3}$ by $HNO_{3}$:
$2HNO_{3}+Na_{2}CO_{3} \rightarrow 2NaNO_{3}+CO_{2}+H_{2}O$
We can observe that the amount of nitric acid required for the complete neutralization is twice the amount of sodium carbonate (according to the reaction's coefficients).
$n(HNO_{3})=2n(Na_{2}CO_{3})=2\frac{m(Na_{2}CO_{3})}{M(Na_{2}CO_{3})}=2\times\frac{0.2040g}{106\frac{g}{mol}}=3.849\times 10^{-3}mol$
Now, when we have calculated the amount of nitric acid required for the complete neutralization, we can obtain the required concentration of the solution by using its volume:
$c(HNO_{3})=\frac{n(HNO_{3})}{V(HNO_{3})}=\frac{3.849\times 10^{-3}mol}{0.01955dm^{3}}=0.1969 \frac{mol}{dm^{3}}$
