Question

$1.2 \,mL$ of acetic acid having density $1.06 \,g \,cm ^{-3}$ 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 $\left(K_{f}\right.$ of water $=1.86 \,K \,kg\, mol ^{-1}$ )

• 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 $=1.2 \times 1.06=1.272 \,g$

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

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

$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} $

$i=1.04$