Astudent 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  :

∵ 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:

∵ 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 %.

the answer is the

2b2t

explanation:

2b2t