Question

The density of 2 M solution of acetic acid (Molecular wt. 60) is 1.02 g $m{l}^{-1}$. The molality of the solution:

• A. $1.0$
• B. $2.0$
• C. $2.22$
• D. $2.25$

Answer 1

The correct option is C $2.22$

$1ml$ of solution contains $1\cdot 02gm$.

$1000ml$ or $1L$ of solution $⟶1\cdot 02\times 1000gm$

$=1020gm$

Number of moles $=Molarity\times V$

$=2\times 1=2moles$

Weight of solute $=Molecularweight\times moles$

$=60\times 2=120gm$

Weight of solvent $=1020-120=900gm$

$Molality=\frac{\text{Moles of solute}}{\text{Weight of solvent ( in Kg)}}$

$Molality=2\times \frac{1000}{900}$

$=2\cdot 22m$

Option C is correct.