Using the following equation:
[tex]M=\frac{n}{L}[/tex]Where M represents the concentration (Molarity), n represents the number of moles of solute and L represents the liters of solution.
If we replace the values of the problem:
[tex]0.33M=\frac{n}{2.3L}\to n=0.33M\cdot2.3L\to n=0.759molesMgCl_2[/tex]To find the number of grams, what we do is to multiply by the molar mass of MgCl2, which is 95.211g:
[tex]0.759molesMgCl_2\cdot\frac{^{}95.211gMgCl_2}{1molesMgCl_2}=72.26gMgCl_2[/tex]Therefore, 72.26gMgCl2 are needed.
