# What volume (in L) of a 2.46 M magnesium nitrate Mg(NO_3)_2 solution would be needed to make 275 mL of a 0.758 M solution by dilution?

May 2, 2016

Under $0.1 \cdot L$

#### Explanation:

We need $275 \cdot m L$ of a solution that is $0.758 \cdot m o l \cdot {L}^{-} 1$ with respect to magnesium nitrate.

$\text{Moles of solute} = 0.275 \cdot L \times 0.758 \cdot m o l \cdot {L}^{-} 1$ $=$ $0.209 \cdot m o l$.

So we divide this molar quantity, by the molarity of the mother solution to gives us the volume:

$\frac{0.209 \cdot m o l}{2.46 \cdot m o l \cdot {L}^{-} 1}$ $=$ $85 \cdot m L$

So we would take this volume and dilute it with fresh solvent to give the required volume. Capisce?