# Calculate the volume in milliliters of 2.35 M potassium hydroxide that contains 13.5 g of solute?

Jun 30, 2017

$\text{102.4 mL}$ of the solution contains $\text{13.5 g}$ of $\text{KOH}$, having a concentration of $\text{2.35 mols KOH"/"L solution}$.

Once you work out the keywords, the math isn't too bad. First, let's define some terms:

• $\text{Molarity" = "mols solute"/"L solution" = "M" = "mols"/"L}$
• $\text{Solute" = "KOH" -= "potassium hydroxide}$
• $\text{Solvent" = "H"_2"O}$ (water)

You are asked to get the $\text{mL}$ (milliliters) of solution that contains $\text{13.5 g}$ of $\text{KOH}$, and you are given the molarity, $\text{2.35 mols/L}$.

So, this is asking you to use the molarity and do a unit conversion. Since you want the volume for your answer, you would start with the orientation of the fraction so that the volume starts on top.

$\text{L"/(2.35 cancel"mols") xx cancel"1 mol KOH"/(56.1056 cancel"g KOH") xx 13.5 cancel"g KOH}$

$=$ $\text{0.1024 L solution}$

But since you were asked for this in $\text{mL}$, there is one more step. There are 1000 millithings in one thing. So:

$\textcolor{b l u e}{{V}_{s o l n}} = 0.1024 \cancel{\text{L" xx "1000 mL"/cancel"L}}$

$=$ $\textcolor{b l u e}{\text{102.4 mL solution}}$

For perspective, in real life, a $\text{mL}$ (a cubic centimeter of water) is 1000 times smaller than $\text{1 L}$. If we have $\text{102.4 mL}$, that's a pretty ordinary amount in the laboratory (we normally have $\text{100.00 mL}$, $\text{200.00 mL}$, and $\text{250.00 mL}$ volumetric flasks available).