# What volume of a 0.40 M Mg(NO_3)_2 solution contains 95 g of Mg(NO_3)_2?

Approx. $1.60 \cdot L$
$\text{Concentration}$ $=$ $\text{Moles of solute"/"Volume of solution}$
And thus $\text{Volume of solution}$ $=$ $\text{Moles of solute"/"Concentration}$
$=$ ((95*g)/(148.32*g*mol^-1))/(0.40*mol*L^-1)~=??L