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Galina-37 [17]
3 years ago
5

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 follow

ing:
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
Chemistry
1 answer:
AleksandrR [38]3 years ago
4 0

Answer:

a.) 86.01 g.

b.) 3.36 g.

c.) 0.394 m ≅ 0.40 m.

d.) 4.77%.

e.) 3.9%.

Explanation:

<em>a.) mass of the solution:</em>

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

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

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

<em>b.) grams of sodium bromide  :</em>

  • 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.

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

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

<em>c.) molality of the solution:</em>

  • 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.

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

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

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

<em>e.) % (m/m) of the solution:</em>

∵ (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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