Question

What is the molality of a 0.735 M aqueous ZnCl2 solution with a density of 1.102 g solution/mL solution. Assume the density of pure water is 1.00 g/mL. The molar mass of ZnCl2 is 136.29 g/mol.

Answer 1

Let the volume of solution be 1 L

Molarity = Number of moles of solute/Volume of solution in L

0.735M = Number of moles of solute/1L

Number of moles of solute = 0.735 moles

Molality = Number of moles of solute/mass of solution in Kg

Mass of solution in Kg = Volume of solution * density of solution

=> 1000 mL * 1.102 g/mL

=> 1102 grams = 1.102 Kg

mass of solvent = mass of solution - mass of solute

=> 1102g - 0.735 mol * 136.29 gm/mol

=> 1102 - 100.17125

=> 1001.8285 grams = 1.0018285 Kg

molality = number of moles of solute/mass of solvent in Kg

=> 0.735/1.0018285

=> 0.7336 m