# How many grams of solute are present in 935 mL of 0.720 M KBr?

Jul 10, 2016

$\text{Grams of KCl}$ $\cong$ $50 \cdot g$

#### Explanation:

$\text{Concentration"="Moles of solute"/"Volume of solution}$. This gives units of $m o l \cdot {L}^{-} 1$.

Thus $\text{moles of solute"="volume"xx"concentration} . =$

$935 \cdot \cancel{m L} \times {10}^{-} 3 \cancel{L} \cdot \cancel{m {L}^{-} 1} \times 0.729 \cdot \cancel{m o l} \cdot \cancel{{L}^{-} 1} \times 74.55 \cdot g \cdot \cancel{m o {l}^{-} 1}$

$=$ ??g