Question

Phenol associates in benzene to certain extent to form a dimer. A solution that is containing 20 x 10-3kg of phenol dissolved in 1 kg of benzene has its freezing point decreased by 0.61 K. Calculate the fraction of phenol that has dimerised.

Answer 1

Answer:

90% of phenol has been dimerised.

Explanation:

Depression in freezing point = 0.61 K

Mass of phenol = $20\times 10^{-3} kg$

Mass of benzeen = =1 kg

$\Delta T_f=i\times K_f\times \frac{20\times g}{94 g/mol\times 1kg}[/ex]</p><p>[tex]0.61 K=i\times 5.12 K kg/ mol \frac{20\times g}{94 g/mol\times 1kg}$

i = 0.56

for association:

$i=1-(1-\frac{1}{n})\alpha$

n = 2 (given)

$\alpha =0.90=90\%$

90% of phenol has been dimerised.

Answer 2

Answer : 90% of phenol has been dimerised.

Explanation :

Depression in freezing point = 0.61 K

Mass of phenol = $20\times 10^{-3} kg$

Mass of benzene = 1 kg

Formula used :

$\Delta T_f=i\times K_f\times m\\\\\Delta T_f=i\times K_f\times\frac{\text{Mass of phenol}}{\text{Molar mass of phenol}\times \text{Mass of benzene in Kg}}$

where,

$\Delta T_f$ = change in freezing point = 0.61 K

i = Van't Hoff factor = ?

$K_f$ = freezing point constant = 5.12 K.kg/mole

m = molality

Now put all the given values in this formula, we get

$\Delta T_f=i\times K_f\times \frac{20\times g}{94 g/mol\times 1kg}\\\\0.61 K=i\times (5.12Kkg/mol)\times \frac{20\times g}{94g/mol\times 1kg}$

i = 0.56

For association :

$i=1-(1-\frac{1}{n})\alpha$

n = 2 (given)

$\alpha=0.90=90\%$

Hence, 90% of phenol has been dimerised.