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18 September, 08:40

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

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Answers (1)
  1. 18 September, 10:31
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    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 %.
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