Challenge problem: an aqueous solution contains 167 g cuso4 in 820 ml of solution. the density of the solution is 1.195 g/ml. calculate the following: a. molarity b. percent by mass c. mole fraction d. molality of the solution

Answers:

a. 1.28 mol/L; b. 17.0 %; c. 0.0227; d. 1.29 mol/kg

Explanation:

a. Molar concentration

c = moles/litres

Moles = 167 × 1/159.61     Do the calculation

Moles = 1.046 mol

Litres = 820 × 1/1000      Do the calculation

Litres = 0.8200 L            Calculate the molar concentration

c = 1.046/0.8200            Do the division

c = 1.28 mol·L⁻¹

b. Percent by mass

Mass % = mass of solute /mass of solution × 100 %

Mass of solution = volume × density     Insert the values

Mass of solution = 820 × 1.195               Do the multiplication

Mass of solution = 979.9 g                     Calculate the mass %

Mass % = 167/979.9 × 100                      Do the operations

Mass % = 17.0 %

c. Mole fraction

χ = moles of solute/(moles of solvent + moles of solute)

Mass of solvent = mass of solution – mass of solute

Mass of solvent = 979.9 – 167         Do the subtraction

Mass of solvent = 812.9 g                Convert to moles

Moles of water = 812.9 × 1/18.02     Do the calculation

Moles of water = 45.11 mol              Calculate the total moles

Total moles = 1.046 + 45.11              Do the addition

Total moles = 46.16 mol                  Calculate the mole fraction

χ = 1.046/46.16                                  Do the division

χ = 0.0227

d. Molal concentration

b = moles of solute/kilograms of solvent

Mass of solvent = 812.9 g = 0.8129 kg     Calculate the molal concentration

b = 1.046/0.8129                                         Do the division

b = 1.29 mol/kg

From the calculations, the final molality of the solution is  1.06 m.

Mass of the solvent can be obtained from;

Mass = density × volume

Mass = 1.195 g/mL ×  820 mL =979.9 g

a) molarity of the solution = number of moles of solute/volume of solution

= 167 g/160 g/mol × 1000/820  = 1.27 M

b) percent by mass of the solute = 167 g/(167 g + 979.9 g) ×100/1

= 14.6%

c) Number of moles of water = 979.9 g/18 g/mol = 54 .4 moles

mole fraction = 1.04 moles/(1.04 moles +  54 .4 moles)

= 0.019

d) molality of the solution = number of moles of solute/mass of solvent in Kg

= 1.04 moles/979.9 ×10^-3 Kg

= 1.06 m

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d

explanation: