# What mass of solute exists in a 1*L volume of 0.650*mol*L^-1 CaBr_2?

Jan 5, 2017

Approx. $130 \cdot g$.

#### Explanation:

$\text{Concentration}$ $=$ $\text{Moles of solute"/"Volume of solution}$, and, clearly, this has units of $m o l \cdot {L}^{-} 1$.

Here we have a $1 L$ volume of $0.650 \cdot m o l \cdot {L}^{-} 1$ $C a B {r}_{2} \left(a q\right)$.

The product,

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

And thus, $1 \cdot \cancel{L} \times 0.650 \cdot m o l \cdot \cancel{{L}^{-} 1} = 0.650 \cdot m o l$ $C a B {r}_{2}$.

And 0.650*cancel(mol)xx199.89*g*cancel(mol^-1)=??*g