In a reaction involving the iodination of acetone, the following volumes were
used to make up the reaction mixture:
$$\text { 10. mL } 4.0 \mathrm{M} \text { acetone }+10 . \mathrm{mL} 1.0
\mathrm{M} \mathrm{HCl}+10 . \mathrm{mL} 0.0050 \mathrm{M} \mathrm{I}_{2}+20 .
\mathrm{mL} \mathrm{H}_{2} \mathrm{O}$$
a. How many moles of acetone were in the reaction mixture? Recall that, for a
component A, moles of \(\mathrm{A}=[\mathrm{A}] \times V,\) where \([\mathrm{A}]\)
is the molarity of \(\mathrm{A}\) and \(V\) is the volume in liters of the
solution of \(\mathrm{A}\) that was used.
_____moles acetone
b. What was the molarity of acetone in the reaction mixture? The volume of the
mixture was \(50 . \mathrm{mL}\), or \(0.050 \mathrm{L},\) and the number of moles
of acetone was found in Part (a). Again,
$$[\mathrm{A}]=\frac{\text { moles of } \mathrm{A}}{V \text { of soln. in
liters }}$$
_____\(\mathrm{M}\) acetone
c. How could you double the molarity of the acetone in the reaction mixture,
keeping the total volume
at \(50 . \mathrm{mL}\) and keeping the same concentrations of \(\mathrm{H}^{+}\)
ion and \(\mathrm{I}_{2}\) as in the original mixture?
