Measures of an acetic acid $(CH_3 COOH)$ solution with molality $\mu=1.00\cdot 10^{-2}$ showed an
elevation of the boiling point of $3.6\cdot 10^{-3}K$
What can we say?
a) $a$ = acetic acid
Each mole gives by dissociation $i$ moles of particles !
$i$ Van t'Hoff´s factor (see discussion of $i$ → here)
$\Delta T$ $=$ $K_{eb}\cdot i\cdot \mu_s$
$i=\frac{\Delta T}{K_{eb}\cdot\mu_a}$
$i=\frac{3.6\cdot 10^{-3}}{0.512\cdot 1.00\cdot 10^{-2}}=1.07$
Calculation of the dissociation degree:
$i=1+(2-1)\alpha$
$\alpha=0.07$
Acetic acid should only be dissociated to $7\%$!
