Question

Density of a 2.05 M solution of acetic acid in water is 1.02 g/mL. The molality of the solution is

• A. 3.28 mol Kg–1
• B. 2.28 mol Kg–1
• C. 0.44 mol Kg–1
• D. 1.14 mol Kg–1

Answer 1

Answer: (B)

2.28 mol Kg–1

Answer 2

Molality, m = $\frac { \mathrm { M } } { 1000 \mathrm {~d} - \mathrm { MM } _ { 2 } }$ × 1000

where M = molarity, d = density, M2 = molecular mass

m = $\frac { 2.05 } { 1000 \times 1.02 - 2.05 \times 60 }$ = 2.28 mol kg–1