Question

Density of 2.05 M solution of acetic acid in water is 1.02 g/ml. find molality of same solution

Answer 1

Take a hypothetical sample of 1.000 L: 

(1000 ml) x (1.02 g/ml) = 1020 g 

(2.05 mol/L) x (1.000 L) = 2.05 mol 

(2.05 mol CH3COOH) x (60.05221 g CH3COOH/mol) = 123.1 g CH3COOH 

(1020 g total) - (123.1 g CH3COOH) = 896.9 g H2O 

(2.05 mol) / ( 896.9 g / 1000 g) = 2.29 m

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