Question

$0.6$ ml of glacial acetic acid with density $1.06$ gm $L^{-1}$ is dissolved in $1$ kg water and the solution froze at $-{0.0205}^o$C. Calculate vant Hoff factor and $K_f$ for water is $1.86$ K kg $mo{l}^{-1}$.

• A. 2.103
• B. 2.36
• C. 1.041
• D. None of these

Answer 1

The correct option is C 1.041

Vant Hoff factor is associated with colligative properties and is the ratio of calculated Molar mass to that of Observed Molar mass.

$\Delta T_f=i\times K_f\times m--------\left(1\right)$

where,

i is Vant Hoff factor

$K_f$ is Cryoscopic constant

m is molality

$Density=\frac{mass}{volume}$

Mass of glacial acetic acid = $\frac{1.06\times 0.6}{1000}$

= $0.136\times {10}^{-3}$

from equation (1)

$0.0205=i\times 1.86\times \frac{0.136\times {10}^{-3}}{60\times 1}$

$i=1.041$