# Question #f9cf5

Apr 14, 2017

$2.50 \cdot m o l$ of $C a C {l}_{2}$ are necessary..........................

#### Explanation:

We use the relationship:

$\text{Concentration}$ $=$ $\text{Moles of solute"/"Volume of solution}$.

And thus $\text{Moles of solute"="Concentration"xx"Volume of solution}$

$= 5 \cdot m o l \cdot {L}^{-} 1 \times 500 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1 = 2.50 \cdot m o l$

What is the mass of this molar quantity?