[tex]=166.3g\ Ba(BrO_{3})_{2}\ x\ \frac{1\ mol\ Ba(BrO_{3})_{2}}{393.13g\ Ba(BrO_{3})_{2}}\ x\ \frac{6.022\ x\ 10^{23}\ particles}{1\ mol\ Ba(BrO_{3})_{2}}[/tex]
[tex]=2.547398062\ x\ 10^{23}\ particles[/tex]
[tex]=2.55\ x\ 10^{23}\ particles[/tex]
You are given the mass of barium bromate. You have to find the moles of this substance using the molar mass and then multiply it by avogadro's number to find how many particles are present.
