Question

How many g of barium fluoride are needed to prepare 91.4 mL of a 2.23 M solution of barium fluoride?

Answer 1

volume , V = 91.4 mL

= 9.14*10^-2 L

use:

number of mol,

n = Molarity * Volume

= 2.23*9.14*10^-2

= 0.2038 mol

Molar mass of BaF2,

MM = 1*MM(Ba) + 2*MM(F)

= 1*137.3 + 2*19.0

= 175.3 g/mol

use:

mass of BaF2,

m = number of mol * molar mass

= 0.2038 mol * 1.753*10^2 g/mol

= 35.73 g

Answer: 35.7 g