Question

Calculate the freezing point and the boiling point of each of the following solutions using observed Van’t Hoff factors in Table 17.6.

(a) 0.050 m MgCl₂

(b) 0.050 m FeCl₃

Short answer

Freezing point

a) 0.05 m ${\mathrm{MgCI}}_2$=$-0.{257}^{0}C$

b) 0.05 m ${\mathrm{FeCI}}_3$=$-0.{316}^{0}C$

boiling point

a) 0.05 m ${\mathrm{MgCI}}_2$=$100.{07}^{0}C$

b) 0.05 m ${\mathrm{FeCI}}_3$= $100.{08}^{0}C$

Step-by-step solution

Subpart a) Step 1:

Freezing point depression is given by the formula

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

and boiling point elevation is given by the formula

$\Delta T_b=i\times m\times K_b$

i = 2.7 [ given in Table 17.6]

m = 0.05 m [ given]

$K_f$=$1.86\frac{{}^{0}C}{m}$

$K_b=0.512\frac{{}^{0}C}{m}$

Putting these values in the equation of freezing point depression an boiling

point elevation we get

$\Delta T_f=2.76\times 0.05\times 1.86=0.{257}^{0}C$

So, the freezing point of this solution will be $0^{0}C-0.{257}^{0}C=-0.{257}^{0}C$

and $\Delta T_f=2.76\times 0.05\times 0.512=0.{070}^{0}C$.

So, the boiling point of this solution will be ${100}^{0}C+0.{070}^{0}C=100.{07}^{0}C$

Step 2:

Now for the solution of 0.05 m FeCl

i = 3.4 [ given in Table 17.6]

m = 0.05 m [ given]

$K_f$=$1.86\frac{{}^{0}C}{m}$

$K_b=0.512\frac{{}^{0}C}{m}$

Putting these values in the equation of freezing point depression an boiling

point elevation we get

$\Delta T_f=3.4\times 0.05\times 1.86=0.{316}^{0}C$.

So, the freezing point of this solution will be $0^{0}C-0.{316}^{0}C=-0.{316}^{0}C$

and $\Delta T_f=3.4\times 0.05\times 0.512=0.{0870}^{0}C$.

So, the boiling point of this solution will be ${100}^{0}C+0.{08}^{0}C=100.{08}^{0}C$