20 % of acetic acid is dissociated when its 5g is added to 500mL of...

Van't Hoff factor (i) for dissociation of acetic acid can be written as follows,

$i = 1 + ( n - 1 ) \alpha$

$n =$Number of ions

$\alpha =$Degree of dissociation

$( i = 1 + 0 . 2 ( 2 - 1 ) = 1 . 2$

${\Delta T}_{f} = {iK}_{f}m{\Delta T}_{f} = Depressioninfreezingpoint{K}_{f} = Molardepressionconstantm = molality$

${\Delta T}_{f} = 1 . 2 \times 1 . 86 \times \dfrac{5 \times 1000}{60 \times 500}$

${\Delta T}_{f} = 0 . 372$

$\Delta{T}_{f} = 372 \times {10}^{- 3}$