# How many mL of a 0.100 M solution are required have 1.32 x 10^2 moles of solute?

Dec 1, 2016

Rather a large volume, over $1 \cdot {m}^{3}$

#### Explanation:

$\text{Concentration}$ $=$ $\text{Moles of solute"/"Volume of solution}$.

And, typically, $\text{concentration}$ has units of $m o l \cdot {L}^{-} 1$.

Here, clearly, $\text{Volume of solution"="Moles of solute"/"Concentration}$

$= \frac{1.32 \times {10}^{2} \cdot m o l}{0.100 \cdot m o l \cdot {L}^{-} 1} = 1320 \cdot L$