# How many grams of CuF_2 are needed to make a 600 mL solution with a molarity of 0.250 M?

Aug 18, 2016

Approx. $15 \cdot g$

#### Explanation:

$\text{Concentration}$ $=$ $\text{number of moles"/"volume of solution}$

Thus $0.250 \cdot m o l \cdot {L}^{-} 1$ $=$ ${n}_{\text{copper fluoride}} / \left(0.600 \cdot L\right)$

And ${n}_{\text{copper fluoride}} = 0.250 \cdot m o l \cdot \cancel{{L}^{-} 1} \times 0.600 \cdot \cancel{L}$

$=$ $0.150 \cdot m o l \text{ copper fluoride}$

$\text{Mass of cupric fluoride}$ $=$ $0.150 \cdot m o l \times 101.54 \cdot g \cdot m o {l}^{-} 1$ $=$ ??g