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 MgCl=$-{0.257}^{0}C$

b) 0.05 m FeCl=$-{0.316}^{0}C$

boiling point

a.0.05 m MgCl =$1000.{07}^{0}C$

b. 0.05 m FeCl =${100.08}^{0}C$

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 ${100}^{0}C+{0.08}^{0}C={100.08}^{0}C$