Question

a) A sucrose solution is prepared to a final concentration of 0.160 M . Convert this...

a) A sucrose solution is prepared to a final concentration of 0.160 M . Convert this value into terms of g/L, molality, and mass % (molecular weight, MWsucrose = 342.296 g/mol ; density, ρsol′n = 1.02 g/mL ; mass of water, mwat = 965.2 g ). Note that the mass of solute is included in the density of the solution. Express the concentrations in grams per liter, molality, and mass percent to three significant figures separated by commas. Hints Sucrose concentrations = g/L, m, % Express the concentrations in grams per liter, molality, and mass percent to three significant figures separated by commas.

B) Suppose 209.1 mg of PbCl2 was added to 15.0 mL of water in a flask, and the solution was allowed to reach equilibrium at 20.0 oC. Some solute remained at the bottom of the flask after equilibrium, and the solution was filtered to collect the remaining PbCl2, which had a mass of 60.5 mg . What is the solubility of PbCl2 (in g/L)? Express the concentration in grams per liter to three significant figures. Solubility of PbCl2 = g/L

Homework Answers

Answer #1

(a)

(i)

Molarity = 0.160 M

Means that 0.160 mol of sucrose is in 1 L of solution

Mass of sucrose = moles * molar mass = 0.160 * 342.296 = 54.767 g.

Therefore, concentration in g/L = 54.767 g/L

(ii)

density of solution = 1.02 g/mL

Mass of 1000 mL of solution = 1.02 * 1000 = 1020 g.

Mass of solute = 54.767 g.

Mass of solvent = 1020 - 54.767 = 945.2 g. = 0.9452 kg.

Molality = moles / kg of solvent = 0.160 / 0.9452 = 0.1693 mol / kg

(iii)

Mass percentage = mass of sucrose * 100 / mass of solution = 54.767 * 100 / 1020 = 5.37 %

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