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 %.
