Explanation:
The given data is as follows.
Volume of solution = 850.0 ml
Volume percentage of ethyl alcohol = 24%
24%(v/v) solution of alcohol means that 24 ml solute of ethyl alcohol is present per 100 ml of solution.
Hence, volume of alcohol present in 850 ml solution is calculated as follows.
= [tex]850 ml \times \frac{\text{24 ml ethyl alcohol}}{\text{100 ml solution}}[/tex]
= 204 ml ethyl alcohol
Now, we will calculate the number of moles of ethyl alcohol as follows.
No. of moles = [tex]204 ml \times \farc{0.789 g}{ml} \times \frac{1 mol}{46.07 g/mol}[/tex]
= 3.49 mol
Volume of solution 850 ml or 0.850 L (as 1 L = 1000 ml).
Concentration of ethyl alcohol will be calculated as follows.
= [tex]\frac{\text{3.49 mol ethyl alcohol}}{0.850 L}[/tex]
= 4.10 mol/L
= 4.10 M
Therefore, molarity of ethyl alcohol in Listerine is 4.10 M.
