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

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

Given Values

Molarity = 2.05 mol/L

MB = CH3COOH = 2(12)+2(16)+4(1) = 60

Dsol = 1.02 gm/ml

Calculate

Molality = $$\frac{Molarity \times 1000}{\left ( D_{sol} \times 1000 \right )-\left ( Molarity \times M_{B} \right )}$$

Molality = $$\frac{2.05 \times 1000}{\left ( 1.02 \times 1000 \right )-\left ( 2.05 \times 60 \right )}$$

Molality = 2.285 mol/Kg