Example 3.17: Determining the Mass of Solute in a Given Volume of Solution

How many grams of \(\ce{NaCl}\) are contained in 0.250 L of a 5.30 molar solution?

Solution

\(m_{\mathrm{NaCl}}\) = ? ...(goal of this calculation)


\(c_{\mathrm{NaCl}} = 5.30\ \mathrm{M}\)


\(V_{\mathrm{solution}} = 0.250\ \mathrm{L}\)


The volume and concentration of the solution are specified, so the chemical amount (mol) of solute is easily computed as demonstrated in example 3.15:

\(n_{\mathrm{NaCl}} = c_{\mathrm{NaCl}} \cdot V_{\mathrm{solution}}\)

\(\ \ \ =5.30\ \mathrm{M} \cdot 0.250\ \mathrm{L}\)

\(\ \ \ =1.325\ \mathrm{mol}\)


Finally, this chemical amount is used to derive the mass of NaCl:

\(m_{\mathrm{NaCl}} = n_{\mathrm{NaCl}} \cdot M_{\mathrm{\ce{NaCl}}}\)

\(\ \ \ =1.325\ \mathrm{mol} \cdot 58.44\ \frac{\mathrm{g}}{\mathrm{mol}}\)

\(\ \ \ =77.4\ \mathrm{g}\)