Chapter 4 - Chemical Reactions - Example 4-8 - Calculating Molarity from Measured Quantities - Page 124: Practice Example B

0.524 M

Moles of acetic acid= $15.0\,mL\, acetic\,acid\times\frac{1.048\,g\,acetic\,acid}{1\,mL\, acetic\,acid}\times\frac{1\,mol\, acetic\,acid}{60.052\,g\, acetic\,acid}$ $=0.261773\,mol$ Molarity= $\frac{\text{Moles solute}}{\text{Litres solution}}=\frac{0.261773\,mol}{500.0\times10^{-3}\,L}$ $=0.524\,M$