# Determine the amount of substance of calcium chloride used

Determine the amount of substance (in moles) of calcium chloride used in this lab. (Hint: $$0.20\rm~M$$ means $$0.20\rm~moles:1~L$$)

I made $$96$$ milliliters of $$\ce{CaCl2}$$ which would be $$0.096\rm~L$$.
$$\rm0.096~L~of~CaCl_2\times\frac{0.20~mol}{1~L} = 0.0192~mol~of~CaCl_2$$

I was wondering if I solved this right.

$$\mathrm{Molarity}=\frac{\mathrm{moles}}{\mathrm{liters}}$$
\begin{align} 0.20 &= \frac{n}{0.096}\\ n &= \pu{0.20 M} \times \pu{0.096 L} \\&= \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: $$n = \pu{0.019 mol}\,\ce{CaCl2}$$.