# Question 33ba6

Jan 22, 2017

Approx. $15 - 16 \cdot g$.

#### Explanation:

$\text{Concentration}$ $=$ $\text{Moles"/"Volume}$

And thus $\text{moles}$ $=$ $\text{concentration}$ $\times$ $\text{volume}$

$\text{moles}$ $=$ $0.280 \cdot m o l \cdot \cancel{{L}^{-} 1} \times 275 \times {10}^{-} 3 \cdot \cancel{L}$

$= 0.0770 \cdot m o l$

And thus we need a mass of $0.0770 \cdot m o l \times 208.3 \cdot g \cdot m o {l}^{-} 1$

$=$ ??*g#