Question

$1.2mL$ of acetic acid having density $1.06gc{m}^{-3}$ is dissolved in $1$ litre of water. The depression in freezing point observed for this concentration of acid was $0.041^{\circ }C$. The van't Hoff factor of the acid is $(K_f$ of water $=1.86Kkgmo{l}^{-1}$ )

• A. 0.41
• B. 1.04
• C. 0.96
• D. 1.54

Answer 1

Answer: (B)

From depression in freezing point,

$\Delta T_f=i\times K_f\times m...(i)$

$\text{ Molality }(m)=\frac{\text{ Moles of solute }}{\text{ Mass of solvent (in g) }}\times 1000$

Mass = density $\times$ volume $=1.2\times 1.06=1.272g$

$\therefore$ Moles of solute $=\frac{1.272}{60}$

$(\because \text{ Moles }=\frac{\text{ Mass }}{\text{ Molecular mass }})$

$0.041=\frac{i\times 1.86\times 1.272>\times 1000}{60\times 1000}=\frac{60\times 0.041}{1.86\times 1.27}$

$i=1.04$