Density of a 2.05 M solution of acetic acid in water is 102 g/ml. The molality of the solution is:

Let’s consider the volume of solution of acetic acid and water as1000 ml
So, we have
1.02 gm of acetic acid in 1 ml of the solution
⇒ 1020 gm in 1000 ml of the solution
Now,
Number of moles = 2.05M × 1L
                          = 2.05 mol
Mass of solute = n × Mol.wt of Acetic acid
                      = 2.05 × 60
                       = 123 g
Mass of solvent = 1020 – 123
                       = 897 g
Hence, Molality = (123/60) × (1000/897)
                       = 2.28 mol kg-1

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