The first step we have to follow to answer this question is to convert the given mass to moles using CaCO3 molar mass:
[tex]44gCaCO_3\cdot\frac{1molCaCO_3}{100.1gCaCO_3}=0.44molCaCO_3[/tex]
Now, use the following formula to find the volume of water needed to prepare the solution:
[tex]\begin{gathered} molarity=\frac{molesofsolute}{litersofsolution} \\ 1.5M=\frac{0.44mol}{L} \\ L=\frac{0.44mol}{1.5M} \\ L=0.29L \end{gathered}[/tex]
It means that 0.29L are needed.
