In a reaction involving the iodination of acetone, the following volumes were
used to make up the reaction mixture: $$10 \mathrm{mL} 4.0 \mathrm{M} \text {
acetone }+10 \mathrm{mL} 1.0 \mathrm{M} \mathrm{HCl}+10 \mathrm{mL} 0.0050
\mathrm{M} \mathrm{I}_{2}+20 \mathrm{mL} \mathrm{H}_{2} \mathrm{O}$$,
a. How many moles of acetone were in the reaction mixture? Recall that, for a
component \(A\), no. 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
b. What was the molarity of acetone in the reaction mixture? The volume of the
mixture was \(50 \mathrm{mL}\) \(0.050 \mathrm{L},\) and the number of moles of
acetone was found in Part (a). Again, \(M_{A}=\frac{\text { no. moles } A}{V
\text { of soln in liters }}\)
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?
