The volume percentage

Definition

Is called volume percent $(\% Vol)$ the number of liters of pure liquid solute that were used to make $100$ liters of solution.

Exercise 1

We mix $50\; mL$ of water with $50\; ml$ of pure ethanol. Knowing that the volumetric masses of pure ethanol and water are respectively $0.798\frac{g}{mL}$ and $0.998\frac{g}{mL}$, calculate the molarity and percentage (meaning: by mass) of ethanol.

Mass of ethanol = $m_1$ $=$ $50\cdot 0.798$ $=$ $39.9 \; g $ Massof water = $m_2$ $=$ $50\cdot 0.998$ $=$ $49.9 \; g $ Molarity of ethanol = $[C_2H_5OH]$ $=$ $\frac{39.9}{46\cdot 0.09626}$ $=$ $9.07 \frac{mol}{L}$ Percentage of ethanol = $\%_{C_2H_5OH}$ $=$ $\frac{39.9\cdot 100}{39.9+49.9}$ $=$ $44.4$

Exercise 2

Calculate the volume percentage of ethanol in exercise 1 !

Volume percentage of ethanol = $\%Vol_{C_2H_5OH}$ $=$ $ \frac{50\cdot 100}{96.26}$ $=$ $51.9$

Exercise 3

An aqueous solution of $40.04\% $ ethanol has a density equal to $0.94$. Calculate the volume percentage !

$\frac{100}{0.94} \; mL$ contain $ \frac{40.04}{0.798}\: mL $ ethanol (see ex 1) $\%Vol_{C_2H_5OH}$ $=$ $ \frac{40.04\cdot 0.94 \cdot 100}{0.798\cdot 100}$ $=$ $ 47.2$