physical chemistry

Question

A 7.71 M solution of water (solute, mw = 18.01) in isopropyl alcohol (solvent, mw = 60.11) has a density of 0.818 g/cm3.

(i) Calculate the percentage by mass of water in the solution

(ii) Calculate the mole fraction of water in the solution.

(iii) Calculate the concentration of water in the solution in ppm

(iv) Calculate the molality of water in the solution.

Answer 1

7.71 M means 7.71 moles solute / liter solution

(i). % by mass of H2O in solution = g H2O / 100 g solution

Assume 1 liter of solution. 7.71 mols H2O x 18.01 g / mol = 138.86 g H2O

1 L solution = 1000 ml x 0.818 g/ml = 818 g total mass of solution

Thus, mass of solvent (IPA) = 818 g - 138.86 g = 679 g

% by mass of H2O = 138.86 g H2O / 818 g solution (x100) = 16.97% = 17.0% H2O by mass

(ii). mole fraction of H2O = moles H2O / total moles

moles H2O = 7.71 moles

moles IPA = 679 g IPA x 1 mol IPA / 60.1 g = 11.3 mols IPA

Total moles = 7.71 + 11.3 = 19.0 moles

mole fraction of H2O = 7.71 / 19.0 = 0.406

(iii). concentration of H2O in ppm

ppm = mg/L

139 g H2O / L x 1000 mg / g = 139,000 mg / L = 139,000 ppm

(iv). molality of H2O

molality = moles of solute per kg solvent

moles of H2O = 7.71 moles

kg of IPA = 679 g x 1 kg/1000 g = 0.679 kg

7.71 mol / 0.679 kg = 11.4 m = molality of H2O