# What is the volume in mL of a 0.606 M KOH solution containing 0.023 mol of solute?

Nov 23, 2017

We use the quotient....$\text{Concentration"="Moles of solute"/"Volume of solution} '$
And thus $\text{Volume"="Moles of solute"/"Concentration}$
$= \frac{0.023 \cdot \cancel{m o l}}{0.606 \cdot \cancel{m o l} \cdot {L}^{-} 1} = 3.80 \times {10}^{-} 2 \cdot \frac{1}{\frac{1}{L}} = 3.80 \times {10}^{-} 2 \cdot L$
And since there are $1000 \cdot m L \cdot {L}^{-} 1$, we gots a volume of $38 \cdot m L$.