Question

A 12.0 % NaCl_(aq) has a density of 1.05 g/mL. What is the molality of the solution?

Answer 1

PRINCIPLE:

Molality of a solution is the number of moles of the solute present in 1kg or 1000 g of the solvent

Step 1: Calculation of number of moles of the solute ,NaCl

Step 2: Calulation of the mass of the solvent

Step 3 : Calculation of molality of the solution

PROCEDURE:

STEP1: Calculation of number of moles of the solute ,NaCl

A 12.0 % NaCl_(aq) has 12 g NaCl in 100 mL solution

Molar mass of NaCl = 23 +35.5 = 58.8 g/mol

Number of moles of NaCl = Mass of NaCl in gram/Molar mass of NaCl

                                             = 12 g/ 58.8 g/mol = 0.2051 mol

Number of moles of solute, NaCl = 0.2051 mol

STEP 2: Calulation of the mass of the solvent, water

Density of the solution = 1.05 g/mL

That means, the mass of 1mL solution = 1.05 g

Therfore , the mass of 100 mL solution = 1.05 x 100 g = 105 g

Mass of Solution = Mass of solvent + Mass of solute

Mass of solvent = Mass of Solution - Mass of solute

                           = 105 g - 12 g = 93 g = 93 x 10^-3 kg

Mass of solvent, water = 93 g = 93 x 10^-3 kg

STEP 3 : Calculation of molality of the solution

93 x 10^-3 kg of water contains 0.2051 mol of NaCl

Therefore 1kg of water will contain [0.2051 mol x 1kg /93 x 10^-3 kg ] mol of NaCl = 2.2054 mol of NaCl

Molality of the solution = 2.2054 mol / kg