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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.

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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 &= \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: $n = \pu{0.019 mol}\,\ce{CaCl2}$.

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