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Formula from Chapter 3: chemical amount of a solute

You can use this if: there is a solution (homogeneous mixture)

\(n_{\mathrm{solute}}\) \(= 0.513\ \mathrm{mol}\)


\(V_{\mathrm{solution}}\) \(= 0.750\ \mathrm{L}\)


\(c_{\mathrm{solute}}\) \(= \dfrac{n_{\mathrm{solute}}}{V_{\mathrm{solution}}}\)

\(\ \ \ =\dfrac{0.513\ \mathrm{mol}}{0.750\ \mathrm{L}}\)

\(\ \ \ =0.684\ \frac{\mathrm{mol}}{\mathrm{L}}\)