A chemist prepares a solution of iron ( II ) bromide by measuring out 59.5 mg of FeBr2 into a 100mL volumetric flask and filling to the mark with distilled water.

Calculate the molarity of Br anions in the chemist's solution.

Be sure your answer is rounded to the correct number of significant digits and answer must be in mol/L

0.00552 mol/L

Explanation:

M(FeBr2) = 215.65 g/mol

59.5 mg = 0.0595 g FeBr2

0.0595 g * 1 mol/215.65 g = 0.0002759 mol FeBr2

FeBr2 2 Br-

1 mol    2 mol

0.0002759  mol FeBr2 * 2 mol Br/1mol FeBr2 = 0.0005518 mol Br-.

100 mL = 0.1 L

0.0005518 mol Br-/ 0.1 L = 0.005518 mol/L≈ 0.00552 mol/L

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