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
\(m_{\mathrm{NaCl}}\) = ?
\(c_{\mathrm{NaCl}}\) \(= 5.30\ \mathrm{M}\)
\(V_{\mathrm{solution}}\) \(= 0.250\ \mathrm{L}\)
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:
\(m_{\mathrm{NaCl}}\) \(= n_{\mathrm{NaCl}} \cdot M_{\mathrm{\ce{NaCl}}}\)
\(\ \ \ =1.325\ \mathrm{mol} \cdot 58.44\ \frac{\mathrm{g}}{\mathrm{mol}}\)
\(\ \ \ =77.4\ \mathrm{g}\)