Homework Help

An aqueous solution made up of 32.47 g of iron(III) chloride in 100.0 mL of solution...

user profile pic

ashweee09 | eNotes Newbie

Posted April 22, 2013 at 5:08 PM via web

dislike 1 like
  1. An aqueous solution made up of 32.47 g of iron(III) chloride in 100.0 mL of solution has a density of 1.249 g/mL. calculate its concentration in Molarity and also in Molality.

1 Answer | Add Yours

user profile pic

llltkl | College Teacher | (Level 3) Valedictorian

Posted April 23, 2013 at 8:14 AM (Answer #1)

dislike 1 like

Ask one question at a time.

Molarity is defined as the number of moles of solute dissolved in 1000 mL of the solution. Here 100 mL of the solution contains 32.47 g of iron (III) chloride (MW==55.845+3*35.453 = 162.2),

So, 1000 mL contains 324.7 g of iron(III) chloride = 324.7/162.2 moles = 2.0018 or, 2.0 moles (approximately).

Hence concentration of the solution is 2.0 Molar.

Molality is defined as the number of moles of solute dissolved per 1000 g of the solvent.

Density of the solution is 1.249 g/mL. Hence, 100.0 mL = 100.0*1.249 =124.9 g solution contains 32.47 g solute. Mass of solvent = 124.9-32.47 = 92.43 g.

So, 92.43 g solvent is associated with 32.47/162.2 moles of solute (iron (III) chloride)

Or, 1000 g solvent is associated with 32.47*1000/(162.2*92.43) moles of solute = 2.17 moles (approximately).

Hence concentration of the solution is 2.17 Molal.

Sources:

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes