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 

Answer:

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

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