Determine the amount of substance of calcium chloride used
Yes, you did it correctly, though there is a small issue with significant figures if that is a concern.

$$\mathrm{Molarity}=\frac{\mathrm{moles}}{\mathrm{liters}}$$

So for your values:

\begin{align}
0.20 &amp;= \frac{n}{0.096}\\
n &amp;= \pu{0.20 M} \times \pu{0.096 L} \\&amp;= \pu{0.0192 mol}\, \ce{CaCl2}
\end{align}

However, both your molarity and volume have only two significant figures and you're multiplying those values, so the result should only be reported to two significant figures as well: $n = \pu{0.019 mol}\,\ce{CaCl2}$.