# Question 7746e

Jun 21, 2015

You have 13.6 g of beryllium fluoride in that solution.

#### Explanation:

You're dealing with a solution of known molarity of beryllium fluoride, $B e {F}_{2}$.

Molarity is defined as moles of solute, in your case beryllium fluoride, per liters of solution. This means that if you know the molarity of the solution and its volume, you can determine how many moles of solute you have present.

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{B e {F}_{2}} = \text{0.442 M" * 655 * 10^(-3)"L" = "0.2895 moles}$ $B e {F}_{2}$

To get the mass of beryllium fluoride that would contain this many moles, use the compound's molar mass

0.2895cancel("moles") * "47.01 g"/(1cancel("mole")) = "13.609 g"# $B e {F}_{2}$

Rounded to three sig figs, the answer will be

${m}_{B e {F}_{2}} = \textcolor{g r e e n}{\text{13.6 g}}$