Chapter 2.2, Problem 23PP

Interpretation Introduction

Interpretation:

Mass of acetic acid in grams present in 185 mL of vinegar needs to be determined if it is 5.00% acetic acid by mass and has density 1.02 g/mL.

Concept introduction:

The relation between density, mass and volume is as follows:

$d=\frac{m}{V}$

Here, m is mass and V is volume.

Answer to Problem 23PP

In 185 mL of vinegar, 9.45 g of acetic acid is present.

Explanation of Solution

Volume of Vinegar $=185mL$

Density of vinegar $=1.02g/mL$

Since, $Mass=Density\times Volume$

Therefore, by putting the value of density and volume, the mass of vinegar in grams $=189g$

Now, in vinegar, 5% is acetic acid, so

$=189\times \frac{5}{100}=9.45$

Thus, 9.45 g of acetic acid is present in vinegar.

Conclusion

Thus, the mass of acetic acid in 185 mL vinegar is 9.45 g.