In a reaction involving the iodination of acetone, the following volumes were
used to make up the reaction mixture: $$5 \mathrm{mL} 4.0 \mathrm{M} \text {
acetone }+10 \mathrm{mL} 1.0 \mathrm{M} \mathrm{HCl}+10 \mathrm{mL} 0.0050
\mathrm{M} \mathrm{I}_{2}+25 \mathrm{mL} \mathrm{H}_{2} \mathrm{O}$$
a. How many moles of acetone were in the reaction mixture? Recall that, for a
component \(A\), moles \(A=M_{A} \times V,\) where \(M_{A}\) is the molarity of \(A\)
and \(V\) is the volume in liters of the solution of \(A\) that was used.
___________ moles acetone
b. What was the molarity of acetone in the reaction mixture? The volume of the
mixture was \(50 \mathrm{mL}\). 0.050 L, and the number of moles of acctone was
found in Part (a). Again, \(M_{A}=\frac{\text { moles of } A}{V \text { of
soln. in liters }}\) ___________________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?
