# Question #f97c6

Jun 22, 2017

Approx......$80 \cdot m L$

#### Explanation:

We need a molar quantity of $\frac{22.0 \cdot g}{213.9 \cdot g \cdot m o {l}^{-} 1} = 0.103 \cdot m o l$ with respect to $\text{aluminum nitrate}$.

Because $\text{concentration"="moles of solute"/"volume of solution (L)}$, we needs the quotient........

$\text{volume"="moles of solute"/"concentration}$

$= \frac{0.103 \cdot m o l}{1.25 \cdot m o l \cdot {L}^{-} 1 \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1}$

$= 82.3 \cdot m L$, note that $\frac{1}{m {L}^{-} 1} = \frac{1}{\frac{1}{m L}} = m L$ as required.........