Answer
$V(HCl) = 18ml$
Work Step by Step
The following neutralization reaction occurs:
$CaCO_{3} + 2HCl \rightarrow CaCl_{2} + H_{2}O + CO_{2}$
The given mass of $CaCO_{3}$ is equal to $900mg = 0.9g$, which converted to moles equals to $n(CaCO_{3})=\frac{m(CaCO_{3})}{M(CaCO_{3})}=\frac{0.9g}{100\frac{g}{mol}}=0.009mol$.
Considering stoichiometric coefficients in the reaction above, we can conclude that the amount of necessary hydrochloric acid is twice the amount of calcium - carbonate. Therefore, $n(HCl) = 2n(CaCO_{3}) = 0.018mol$. Since the concentration of $HCl$ solution is equal to $1M$, the volume is equal to $V(HCl) = \frac{n(HCl)}{c(HCl)} = \frac{0.018mol}{1\frac{mol}{dm^3}} = 0.018 dm^3 = 18ml$
