A chemist prepares a solution of sodium bromide by measuring out of into a volumetric flask and filling to the mark with distilled water. Calculate the molarity of anions in the chemist's solution. Be sure your answer is rounded to significant digits.

The molarity of the Br anion is 0.00136 M = 0.0014 M to 2 s.f

Explanation:

Complete full question

A chemist prepares a solution of sodium bromide (NaBr) by measuring out 14. mg of NaBr into a 100 mL 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 2 significant digits.

To do this, we first calculate the molarity of the aqueous solution of NaBr.

Molarity = (Concentration in g/L) ÷ (Molar Mass)

(Concentration in g/L)

= (Mass of solute in g) ÷ (Volume of solution in L)

Mass of solute = 14 mg = 0.014 g

Volume = 100 mL = 0.10 L

(Concentration in g/L)

= (Mass of solute in g) ÷ (Volume of solution in L)

(Concentration in g/L) = (0.014/0.1) = 0.14 g/L

Molarity = (Concentration in g/L) ÷ (Molar Mass)

Molar Mass = 102.894 g/mol

Molarity = (0.14/102.894) = 0.0013606236 M = 0.00136 M

Assuming complete dissociation, NaBr dissociates into

NaBr → Na⁺ + Br⁻

1 mole of NaBr gives 1 mole of Br⁻

0.00136 M of NaBr will give 0.00136 M of Br⁻

So, the molarity of the Br anion is 0.00136 M = 0.0014 M to 2 s.f

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