Chapter 15.2, Problem 2RQ

Interpretation Introduction

Interpretation: Molarity of vinegar solution when its density is $1\text{g/ml}$ and molar mass of acetic acid is $60.05\text{g/mol}$ has to be determined.

Concept introduction: Molarity is a concentration term that is defined as the number of moles of solute dissolved in one litre of the solution. It is denoted by M, and its unit is mol/L.

  $\text{Molarity }=\frac{\text{amount}\left(\text{mol}\right)\text{of}\text{solute}}{\text{volume }\left(\text{L}\right)\text{of solution}}$

Answer to Problem 2RQ

Molarity of vinegar solution when its density is $1\text{g/ml}$ and molar mass of acetic acid is $60.05\text{g/mol}$ is $0.835\text{M}$ .

Explanation of Solution

Given information: Density of vinegar solution is $1\text{g/ml}$ , molar mass of acetic acid is $60.05\text{g/mol}$ , mass of acetic acid is $\text{33}\text{g}$ and mass of vinegar solution is $\text{658}\text{g}$ .

Moles of acetic acid can be calculated by formula as follows:

  $\text{Moles of acetic acid}=\frac{\text{mass}\text{of acetic acid}}{\text{molar mass}\text{of acetic acid}}$

Substitute $\text{33}\text{g}$ for mass of acetic acid and $60.05\text{g/mol}$ for molar mass of acetic acid in above equation.

  $\begin{array}{c}\text{Moles of acetic acid}\text{= }\frac{\text{mass}\text{of acetic acid}}{\text{molar mass}\text{of acetic acid}}\\ \text{=}\frac{\text{33}\text{g}}{\text{60}\text{.05}\text{g/mol}}\\ =0.5495\text{mol}\end{array}$

Volume of vinegar solution is calculated by formula as follows:

  $\text{Volume of vinegar solution}=\frac{\text{mass}\text{of vinegar solution}}{\text{density}\text{of vinegar solution}}$

Substitute $\text{658}\text{g}$ for mass of vinegar solution and $1\text{g/ml}$ for density of vinegar solution in above solution.

  $\begin{array}{c}\text{Volume of vinegar solution}=\frac{\text{mass}\text{of vinegar solution}}{\text{density}\text{of vinegar solution}}\\ =\frac{\text{658}\text{g}}{1\text{g/mL}}\\ =\text{658}\text{mL}\end{array}$

Now, $\text{Molarity of vinegar solution}=\frac{\text{moles of acetic acid }}{\text{volume}\text{of vinegar solution}}$

Substitute $0.5495\text{mol}$ for moles of acetic acid and $\text{658}\text{mL}$ for volume of vinegar solution in above equation.

  $\begin{array}{c}\text{Molarity of vinegar solution}=\frac{\text{moles of acetic acid }}{\text{volume}\text{of vinegar solution}}\\ =\frac{0.5495\text{mol}}{\text{658}\text{mL}\left(\frac{1\text{L}}{1000\text{mL}}\right)}\\ =0.835\text{M}\end{array}$

Conclusion

Molarity of vinegar solution when its density is $1\text{g/ml}$ and molar mass of acetic acid is $60.05\text{g/mol}$ is $0.835\text{M}$ .