Question

The molal depression constant for water=1.85 deg/molal
and for benzene is 5.12 deg/molal. If the ratio of the
latent heats of fusion of benzene to water is 3:8, calculate
the freezing point of benzene.

Answer 1

Answer: Freezing point of Benzene = 5.12° C

Explanation:

Freezing point of water = 0° C = 273 K

Molal depression constant can be expressed by the formula,

$K_b$ = $\frac{RT_b^2}{1000L_v}$

Where, R = gas constant (in calorie), $T_b$= boiling point of solvent in Kelvin

$L_v$ = Latent heat of vaporisation (calorie /gram)

From the above formula, we can write

$K_bL_v$ α $T^2$  (Directly Proportional)

$\frac{K_(water)L(water)}{ K_(benzene)L_(benzene)}$  = $\frac{T_ (water)^{2}}{T_ (benzene) ^2}$

$\frac{1.85 X 8}{5.12 X 3}$ = $\frac{273}{T_ (benzene) ^2}$

$T_ (benzene) ^2$  =  $\frac{273^2 X 5.12 X 3}{1.85 X 8}$

$T_{benzene}$ = 273 $\sqrt{\frac{5.12 X 3}{1.85 X 8} }$

$T_{benzene}$ = 278.12 K

$T_{benzene}$  = 5.12° C

Hence, Freezing point of benzene is 5.12° C.

Answer 2

Thus the temperature of the benzene is 5.117 °C

Explanation:

Given data:

• Modal depression constant for water = 1.85 deg / molal
• Modal depression constant for benzene = 5.12 deg / molal
• Ratio of the  latent heats of fusion of benzene to water = 3 : 8

Solution:

Kf = RT^2 / 100 Lf

kf Lf ∝ T^2

K(water) L (water) / K (benzene) L (benzene) = [T (water)]^2 / [T (benzene)]^2

1.85 / 5.12 = (273)^2 /  [T (benzene)]^2

[T (benzene)] = 273 √ 5.12 x 3 / 1.85 x 8

[T (benzene)]  = 5.117 °C

Thus the temperature of the benzene is 5.117 °C