How do you convert from %w/v to molarity?
Answer:
We define \(\% \mathrm{w} / \mathrm{v}\) as:
By unit conversion, we have that \(\mathrm{g} \times \frac{\mathrm{mol}}{\mathrm{g}} \rightarrow \mathrm{mol}\), and that \(\mathrm{mL} \times \frac{1 \mathrm{~L}}{1000 \mathrm{~mL}} \rightarrow \mathrm{L}\).
Define:
- \(m_{\text {solute }}\) for the solute mass in \(g\)
- \(V_{\text {so } \ln }\) for the solution volume in \(\mathrm{mL}\)
- \(M_{\text {solute }}\) for the molar mass of the solute in \(\mathrm{g} / \mathrm{mol}\)
Therefore:
\(\text { Molarity }=\left(\frac{m_{\text {solute }}}{V_{\text {soln }}} \times 100 \%\right) \times \frac{1000 \mathrm{~mL} / \mathrm{L}}{100 \% \times M_{\text {solute }}}\)Or, rewriting in terms of \(\% \mathrm{w} / \mathrm{v}\) and implied unit cancellation, we have:
\(\text { Molarity }\left(\frac{\text { mol }}{\mathbf{L}}\right)=\frac{\mathbf{1 0 0 0}}{\mathbf{M}_{\text {solute }}} \frac{\% \mathbf{w} / \mathbf{v}}{\mathbf{1 0 0 \%}}\)As an example, if we have a \(37 \% \mathrm{w} / \mathrm{v}\) aqueous \(\mathrm{HCl}\) solution, then:
\(\text { Molarity }\left(\frac{\mathrm{mol}}{\mathrm{L}}\right)=\frac{1000}{36.4609} \frac{37 \% \mathrm{w} / \mathrm{v}}{100 \%}\)
\(=10.15 \mathrm{~mol} / \mathrm{L}\)