A student has 70.5 mL of a 0.463 M aqueous solution of sodium bromide. The density of the solution is 1.22 g/mL. Find the following:

a.) mass of the solution

b.) grams of sodium bromide

c.) molality of the solution

d.) % (m/v) of the solution

e.) % (m/m) of the solution

a.) 86.01 g.

b.) 3.36 g.

c.) 0.394 m ≅ 0.40 m.

d.) 4.77%.

e.) 3.9%.

Explanation:

a.) mass of the solution:

The density of the solution is the mass per unit volume.

∵ Density of solution = (mass of solution) / (volume of the solution).

∴ Mass of the solution = (density of solution) * (volume of the solution) = (1.22 g/mL) * (70.5 mL) = 86.01 g.

b.) grams of sodium bromide:

Molarity (M) is defined as the no. of moles of solute dissolved per 1.0 L of the solution.

∵ M = (no. of moles of NaBr) / (Volume of the solution (L))

∴ no. of moles of NaBr = M * (Volume of the solution (L)) = (0.463 M) * (0.0705 L) = 0.0326 mol.

∵ no. of moles of NaBr = (mass of NaBr) / (molar mass of NaBr)

∴ mass of NaBr = (no. of moles of NaBr) * (molar mass of NaBr) = (0.0326 mol) * (102.894 g/mol) = 3.36 g.

c.) molality of the solution:

Molality (m) is defined as the no. of moles of solute dissolved per 1.0 kg of the solvent.

∵ m = (no. of moles of NaBr) / (mass of the soluvent (kg))

no. of moles of NaBr = 0.0326 mol,

mass of solvent = mass of the solution - mass of NaBr = 86.01 g - 3.36 g = 82.65 g = 0.08265 kg.

∴ m = (no. of moles of NaBr) / (mass of the soluvent (kg)) = (0.0326 mol) / (0.08265 kg) = 0.394 m ≅ 0.40 m.

d.) % (m/v) of the solution:

∵ (m/v) % = [ (mass of solute) / (volume of the solution) ] * 100

∴ (m/v) % = [ (3.36 g) / (70.5 mL) ] * 100 = 4.77%.

e.) % (m/m) of the solution:

∵ (m/m) % = [ (mass of solute) / (mass of the solution) ] * 100

∴ (m/m) % = [ (3.36 g) / (86.01 g) ] * 100 = 3.9 %.