Example 3.18: Determining the Volume of Solution Containing a Given Mass of Solute
In
Example 3.
15, we found the typical concentration of vinegar to be
0.839 M. What volume of vinegar contains
75.6 g of acetic acid?
Solution
\(c_{\mathrm{acetic\ acid}}\) \(= 0.839\ \mathrm{M}\)
\(m_{\mathrm{acetic\ acid}}\) \(= 75.6\ \mathrm{g}\)
\(V_{\mathrm{vinegar}}\) = ?
First, use the molar mass to calculate chemical amount of acetic acid
\(\ce{(CH3CO2H)}\) from the given mass:
\(n_{\mathrm{acetic\ acid}}\) \(= \dfrac{m_{\mathrm{acetic\ acid}}}{M_{\mathrm{\ce{CH3CO2H}}}}\)
\(\ \ \ =\dfrac{75.6\ \mathrm{g}}{60.052\ \frac{\mathrm{g}}{\mathrm{mol}}}\)
\(\ \ \ =1.259\ \mathrm{mol}\)
Then, use the concentration of the solution to calculate the volume of solution containing this chemical amount of solute:
\(c_{\mathrm{acetic\ acid}} = \dfrac{n_{\mathrm{acetic\ acid}}}{V_{\mathrm{vinegar}}}\)
\(\ \ \ =\dfrac{1.259\ \mathrm{mol}}{0.839\ \mathrm{M}}\)
\(\ \ \ =1.500\ \mathrm{L}\)
We could have combined these two steps into one to get the same answer:
\(V_{\mathrm{vinegar}} = \dfrac{m_{\mathrm{acetic\ acid}}}{M_{\mathrm{\ce{CH3CO2H}}} \cdot c_{\mathrm{acetic\ acid}}}\)