Density of `2.05 M` solution of acetic acid in water is `1.02g//mL`. The molality of same solution is:

Options

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

A

Explanation

Mass of 1 L solution = `1.02xx1000 = 1020g`
Mass of 2.05 moles of acetic acid
`=2.05xx60=123g`
Mass of water = `1020-123=897g=0.897kg`
Thus molality of solution `=(2.05mol)/(0.297kg)=2.285m`