Example 3.24: Calculations using Volume Percentage
Rubbing alcohol (isopropanol) is usually sold as a 70%vol aqueous solution. If the density of isopropyl alcohol is
0.785 g/mL, how many grams of isopropyl alcohol are present in a
355 mL bottle of rubbing alcohol?
Solution
\(f_{\mathrm{by\ vol}}\) \(= 70\ \mathrm{％}\)
\(V_{\mathrm{solution}}\) \(= 355.\ \mathrm{mL}\)
\(ρ_{\mathrm{iPrOH}}\) \(= 0.785\ \frac{\mathrm{g}}{\mathrm{mL}}\)
\(m_{\mathrm{iPrOH}}\) = ?
Per the definition of volume percentage, the isopropanol volume is 70% of the total solution volume. Multiplying the isopropanol volume by its density yields the requested mass:
\(m_{\mathrm{iPrOH}}\) \(= f_{\mathrm{by\ vol}} \cdot V_{\mathrm{solution}} \cdot ρ_{\mathrm{iPrOH}}\)
\(\ \ \ =70\ \mathrm{％} \cdot 355.\ \mathrm{mL} \cdot 0.785\ \frac{\mathrm{g}}{\mathrm{mL}}\)
\(\ \ \ =248.50\ \mathrm{mL} \cdot 0.785\ \frac{\mathrm{g}}{\mathrm{mL}}\)
\(\ \ \ =195.1\ \mathrm{g}\)