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

Options

A. `1.14 mol kg^(-1)`
B. `3.28 mol kg^(-1)`
C. `2.28 mol kg^(-1)`
D. `0.44 mol kg^(-1)`

Answer

C

Explanation

Molality, `m=M/(1000d - MM_2)xx1000`
where, M=molarity, d= density, `M_2` = molecular mass
`m=(2.05)/(1000 xx 1.02 - 2.05 xx 60)= (2.05)/897 = 2.28 xx 10^(-3) mol g^(-1) = 2.28 mol kg^(-1)`