The density of an aqueous NaCl solution of molality 4.991 mol kg–1 is 1.169022 g cm–3 at 25 °C and 1 bar?

Question

The density of an aqueous NaCl solution of molality 4.991 mol kg–1 is 1.169022 g cm–3 at 25 °C and 1 bar. What is the molar concentration of the solution?

Answer 1

Molality is defined as the number of moles of solute per kg of solvent. (Not, as the answerer hahaha has assumed, # of moles of solute per kh of *solution*. His/her answer is incorrect.). In this case, we have 4.991 moles of NaCl per 1 kg of water.


Convert this to the mass of NaCl per kg of water using the molecular mass of NaCl (58.4428 g/mol):

4.991 moles NaCl * 58.4428 g NaCl/mol NaCl = 291.688 gm NaCl

The total mass of the solution containing 1 kg of solvent is then the mass of the solvent plus the mass of the NaCl:

mass solution = 291.688 gm + 1000 gm = 1291.688 gm

The volume of this mass of solution is given by the mass divided by the density:

volume solution = (1291.688 gm)/(1.169022 gm/cm^3)

volume solution = 1.105 *10^3 cm^3 = 1.105 liters

The molarity of the solution is the number of moles of NaCl in this volume of solution divided by the volume of the solution:

molarity = (4.991 mol)/(1.105 liters) = 4.517 molar

Answer 2

Take 1 kg of this solution for consideration.
it contains 4.991 mol of NaCl, and it has a volume of:
(1000g)/(1.169022 g cm–3) = 855.4 ml
Let the molar concentration of the solution to be X. We have a simple ratio: X mol/1000ml = 4.991 mol/ 855.4 ml
Thus the molar concentration of the solution is:
(1000ml/855.4 ml)*4.991 mol = 5.835 mol ==> 5.835 M