We can do the calculations one at a time. First we calculate the original concentration.
\begin{align*}
\frac{10 \text{ mL}}{10+90 \text{ mL}} \amp = \frac{10}{100}\\
\amp = \frac{1}{10}.
\end{align*}
To dilute to a ratio of 1/2 we can calculate the amount of diluent to add as a proportion problem like in
Example 2.4.1.
\begin{align*}
\frac{1}{2} \amp = \frac{100 \text{ mL}}{T \text{ mL}}\\
\frac{2}{1} \amp = \frac{T \text{ mL}}{100 \text{ mL}}\\
200 \text{ mL} \amp = T.
\end{align*}
The total will be 200 mL so we need to add \(200 \text{ mL}- 100 \text{ mL} = 100 \text{ mL}\) of additional diluent. Note at this point the concentration is
\begin{equation*}
\frac{10 \text{ mL acyl chloride}}{200 \text{ mL diluent}} = \frac{1}{20}.
\end{equation*}
To dilute again to a ratio of 1/3 we can calculate the amount of diluent to add
\begin{align*}
\frac{1}{3} \amp = \frac{200 \text{ mL}}{T \text{ mL}}\\
1 \cdot (T \text{ mL}) \amp = 3 \cdot (200 \text{ mL})\\
T \amp = 600 \text{ mL}.
\end{align*}
The total will be 600 mL so we need to add \(600 \text{ mL} - 200 \text{ mL} = 400 \text{ mL}\) of additional diluent. Note at this point the concentration is
\begin{equation*}
\frac{10 \text{ mL acyl chloride}}{600 \text{ mL diluent}} = \frac{1}{60}.
\end{equation*}
Now we can determine what the resulting dilution ratio after diluting twice (1/2 and then 1/3).
\begin{align*}
\frac{1}{10} \cdot F \amp = \frac{1}{60}\\
10 \cdot \frac{1}{10} \cdot F \amp = 10 \cdot \frac{1}{60}.\\
F \amp = \frac{1}{6}.
\end{align*}
Notice that
\(\frac{1}{6} = \frac{1}{2} \cdot \frac{1}{3}\text{,}\) that is, the resulting dilution factor is the product of the serial dilutions. This relationship is always true for serial dilution.
