# How many moles of CaCl_2 are in 250 mL of a 3.0 M of CaCl_2 solution?

May 7, 2016

$0.75$mol

#### Explanation:

$C = \frac{n \setminus m o l}{V}$, where $C$ is concentration, $n$ is number of moles and $V$ is volume in liters.

In this problem, $C = 3.0 M$ and $V = 0.25 L$

Substitute this into the equation.

$3.0 M = \frac{n \setminus m o l}{0.25 L}$

$3.0 \left(\frac{m o l}{\cancel{L}}\right) \cdot 0.25 \cancel{L} = 0.75 \setminus m o l$