Example 3.15: Deriving Moles and Volumes from Molar Concentrations
How much sugar (chemical amount) is contained in a modest sip (~10 mL) of the soft drink from
Example 13?
Solution
\(V_{\mathrm{sip}}\) \(= 10\ \mathrm{mL}\)
\(n_{\mathrm{sugar}}\) = ?
We derived derived the concentration of sugar,
0.375 M, in
Example 13,
0.375 M:
\(c_{\mathrm{sugar}}\) \(= 0.375\ \mathrm{M}\)
In this case, we can rearrange the definition of concentration to isolate the quantity sought,
\(n_{\mathrm{sugar}}\), the chemical amount of sugar.
\(c_{\mathrm{sugar}} = \dfrac{n_{\mathrm{sugar}}}{V_{\mathrm{sip}}}\)
\(\ \ \ =0.375\ \mathrm{M} \cdot 10\ \mathrm{mL}\)
\(\ \ \ =3.75\times 10^{-3}\ \mathrm{mol}\)
Or expressed in millimol:
\(n_{\mathrm{sugar}}\) \(= 3.75\times 10^{-3}\ \mathrm{mol}\)
\(\ \ \ =3.75\ \mathrm{mmol}\)