Question

The density of 2.0 M solution of acetic acid in water is 1.2 g/ ml . Calculate the molality of the solution.

Answer 1

Answer:

Suppose the volume of solution is 1000 ml.

1.02 gm of acetic acid in 1 ml of soln.

1020 gm in 1000 ml of soln.

Number ofr moles =2⋅05M×1 lit=2.05 mol

Mass of solute =n×M⋅wt=2.05×60=123 gm

Mass of solvent =1020−123=897 gm

Molality =

60

123

×

897

1000

=2.28 mol kg

−1

Answer 2

Given:-

→ Molarity of the solution = 2.0 M

→ Density of the solution = 1.2 g/mL

To find:-

→ Molality of the solution.

Solution:-

It is clear that acetic acid is solute and water is solvent .

• Molar mass of Carbon (C) = 12 g/mol

• Molar mass of Hydrogen (H) = 1 g/mol

• Molar mass of Oxygen (O) = 16 g/mol

∴ Molar mass of acetic acid [CH₃COOH] :-

= 12 + 1×3 + 12 + 16 + 16 + 1

= 57 + 3

= 60 g/mol

Mass of acetic acid in 1 L solution :-

= 2 × 60

= 120g

Mass of 1 L solution :-

= 1000 × density

= 1000 × 1.2

= 1200g

Mass of water [solvent] in solution :-

= 1200 - 120

= 1080g

Molality of a solution:-

= Moles of solute/Mass of solvent in kg

= 2/[1080/1000]

= 2/1.08

= 1.85 m

Thus, molality of the solution is 1.85m .