Respuesta :
Answer:
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
Hope this Helps!!!
