Question

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

• A. $1.14molk{g}^{-1}$
• B. $3.28molk{g}^{-1}$
• C. $2.28molk{g}^{-1}$
• D. $0.44molk{g}^{-1}$

Answer 1

The correct option is C $2.28molk{g}^{-1}$

2.05 M solution means 1 L of the solution has 2.05 moles of acetic acid.

The molecular weight of acetic acid is 60 g/mol.

The mass of acetic acid $=60g/mol\times 2.05mol=123g$.

The density of solution is 1.02 g/mL

The volume of solution is 1 L or 1000 mL.

Mass of solution is $1.02g/mL\times 1000mL=1020g$

Mass of water $=1020g-123g=897g$

$=1020g-123g=897g=0.897kg$

Molality, $m=\frac{2.05mol}{0.897kg}=2.28mol/kg$

Hence, option $C$ is correct.