Question

Density of a 2.05M solution of acetic acid in water is 1.02 g/mL. The molality of the solution is(a) 2.28 mol kg⁻¹(b) 0.44 mol kg⁻¹(c) 1.14 mol kg⁻¹ d) 3.28 mol kg⁻¹

Answer 1

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   ----  (a)