How many fluoride ions are present in 65.5 ml of 0.210 m alf3 solution?

Hey there!:

Volume in liters of solution:

65.5 mL / 1000 => 0.0655 L

Number of moles of fluoride ions :

AlF3 = 3 atoms of F so :

n = M * V

n = 0.210  * 3 *  0.0655

n = 0.041265 moles of  fluoride ions

Therefore:

Number of ions  :

0.041265* ( 6.02*10²³ ) =  2.485 *10²² ions  

Hope that helps!

the answer is barium and copper bromide

50.23 i believe is correct but don't quote me on that.