# How many grams of solute are present in 705 mL of 0.900 M KBr?

May 14, 2018

Approx. $75 \cdot g$....

#### Explanation:

By definition....$\text{concentration"-="moles of solute"/"volume of solution}$

And so to get the $\text{moles of solute}$, we take the product....

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

And so here we gots.......

${n}_{\text{KBr}} = 0.900 \cdot m o l \cdot {L}^{-} 1 \times 705 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1$

$= 0.6345 \cdot m o l$...

And thus a MASS of 0.6345*cancel(mol)xx119.0*g*cancel(mol^-1)=??*g