# How do you convert from %"w/v" to molarity?

##### 1 Answer
May 15, 2017

We define %"w/v" as:

%"w/v" = "solute mass in g"/"solution volume in mL" xx 100%

By unit conversion, we have that $\text{g" xx "mol"/"g" -> "mol}$, and that $\text{mL" xx "1 L"/"1000 mL" -> "L}$. Define:

• ${m}_{s o l u t e}$ for the solute mass in $\text{g}$
• ${V}_{s o \ln}$ for the solution volume in $\text{mL}$
• ${M}_{s o l u t e}$ for the molar mass of the solute in $\text{g/mol}$

Therefore:

"Molarity" = (m_(solute)/V_(soln) xx 100%) xx (1000 "mL/L")/(100% xx M_(solute))

Or, rewriting in terms of %"w/v" and implied unit cancellation, we have:

bb("Molarity" ("mol"/"L") = 1000/(M_(solute))(%"w/v")/(100%))

As an example, if we have a 37% $\text{w/v}$ aqueous $\text{HCl}$ solution, then:

color(blue)("Molarity" ("mol"/"L")) = 1000/(36.4609) (37cancel(%) "w/v")/(100cancel(%))

$=$ $\textcolor{b l u e}{\text{10.15 mol/L}}$