Question

Calculate the freezing point (in degrees C) of a solution made by dissolving 4.16 g of anthracene {C14H10} in 66.3 g of cyclohexane. The Kfp of the solvent is 20.1 K/m and the normal freezing point is 6.5 degrees C.

Answer 1

Answer:

The expected freezing point of a solution of anthracene in cyclohexane is -0.585°C.

Explanation:

$\Delta T_f=T-T_f$

$\Delta T_f=i\times K_f\times m$

where,

$T_f$ = Freezing point of solution

T = Freezing point of pure solvent

$\Delta T_f$ =depression in freezing point =  

i = van't Hoff factor

$K_f$ = freezing point constant  of solvent

m = molality

We have :

i = 1 ( non electrolyte)

Freezing point constant  of cyclohexane = $K_f$ =1.86°C/m ,

Mass of solvent (cyclohexane) = 66.3 g = 0.0663 kg( 1 g = 0.001 kg)

Mass of solute (anthracene) = 4.16 g

Moles of anthracene = $\frac{4.16 g}{178 g/mol}=0.02337 mol$

$Molality(m)=\frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}$

$m = \frac{0.02337 mol}{0.0663 kg}=0.3525 m$

$\Delta T_f=1\times 20.1^oC/m\times 0.3525 m$

$\Delta T_f=7.085^oC$

Freezing point of pure cyclohexane = T =  6.5°C

Freezing point of solution = $T_f$

$\Delta T_f=T-T_f$

$T_f=T-\Delta T_f=6.5^oC-7.085^oC=-0.585 ^oC$

The expected freezing point of a solution of anthracene in cyclohexane is -0.585°C.