Question

The density of a 2.05M solution of acetic acid in water is 1.02 g/ml. What is its molality?

Answer 1

Answer:

Suppose the volume of solution is 1000 ml

1 ml=1.2 gm

1000 ml=1020 gm

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

Mass of solute =n×M⋅wt=2.05×60=123 gm

Solvent mass =1020−123=897 gm

Molality =

$\frac{123}{60} \times \frac{1000}{897}$

$2.28mol {kg}^{ - 1}$

Answer 2

Answer:

  m = 2.285mol kg⁻¹

Explanation:

molality - m

molarity - M

Density - d

M₂ - Molecular mass - CH₃COOH = 60g

m = 1000M / 1000d - MM₂

m = 1000 × 2.05 / (1000 × 1.02) - (2.05 × 60)

m = 2050 / 1020 - 123

m = 2050 / 897

m = 2.285mol kg⁻¹