Question

The density of a 2.05 M solution of acetic acid in water is 1.02 g/ml. The molality of the solution is:

Answer 1

Part 1:

Molality:

• It is denoted by m.
• It is defined as the number of moles of solute dissolved in per kg of the solvent.

Mathematically:

$\mathrm{m}=\frac{{\mathrm{W}}_{\mathrm{B}}\times 1000}{{\mathrm{M}}_{\mathrm{B}}\times {\mathrm{W}}_{\mathrm{A}}}$

Where,

m = molality

W_B = weight of solute

M_{B =} molecular mass of solute

W_A = Weight of the solvent

Part 2:

Let’s consider the volume of the solution of acetic acid (${\mathrm{CH}}_3\mathrm{COOH}$) and water as $1000$ 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 = $2.05\times 1\text{L}$
= $2.05\text{mol}$
Mass of solute = n × Molecular weight of acetic acid.
= $$\begin{gathered} 2.05\times 60 \\ =123\text{gm} \end{gathered}$$
Mass of solvent = $1020-123$
= $897\text{gm}$
Molality (m) = $\frac{2.05\times 1000}{897}$
= $2.28{\text{mol Kg}}^{-1}$

Therefore, the molality of the solution is $2.28{\text{mol Kg}}^{-1}$