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 \(V\) the volume in liters of the solution of \(A\) that was used.
__________________moles acetone. What was the molarity of acetone in the
reaction mixture? The volume of the mixture was \(50 \mathrm{mL}\), 0.050 liter,
and the number of moles of acetone was found in Part a. Again,
\(M_{A}=\frac{\text { no. moles } A}{V \text { of soln. in liters }}\)
__________________M acetone. 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?
