Question

Evaluate the freezing point of${\mathrm{MgCl}}_2$ if equimolal solutions of $\mathrm{NaCl}$ and ${\mathrm{MgCl}}_2$ are prepared in water. The freezing point of $\mathrm{NaCl}$ is found to be $-2^{\circ }C.$

Answer 1

Freezing point:

• Freezing point depression is a colligative property.
• Freezing point refers to the lowering of the freezing point of solvents upon the addition of non-volatile solutes.
• The freezing point is proportional to the molality of the added solute.
• Mathematically, depression in freezing point can be written as

${\mathrm{\Delta T}}_{\mathrm{f}}=\mathrm{i}\times {\mathrm{K}}_{\mathrm{f}}\times \mathrm{m}$

Where,

• $\Delta T_f$ is the depression in freezing point,
• i is the Van’t Hoff factor,
• $K_f$ is the cryoscopic constant,
• m is the molality.

Step 1: Calculating Van'f Hoff factor

For $\mathrm{NaCl}$

The dissociation of sodium chloride can be represented as:

$\underset{\mathrm{Sodium}\mathrm{chloride}}{\mathrm{NaCl}\left(\mathrm{aq}\right)}\to \underset{\mathrm{Sodium}\mathrm{ion}}{{\mathrm{Na}}^{+}\left(\mathrm{aq}\right)}+\underset{\mathrm{Chloride}\mathrm{ion}}{{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)}$

Now, i = 2

For ${\mathrm{MgCl}}_2$

The dissociation of Magnesium chloride can be represented as:

$\underset{\mathrm{Magnesium}\mathrm{chloride}}{{\mathrm{MgCl}}_2\left(\mathrm{aq}\right)}\to \underset{\mathrm{Magnesium}\mathrm{ion}}{{\mathrm{Mg}}^{2+}\left(\mathrm{aq}\right)}+\underset{\mathrm{Chloride}\mathrm{ion}}{2{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)}$

i = 3

Step 2: Calculating molality

$K_f$ for ${\mathrm{H}}_2\mathrm{O}$ =$1.86\mathrm{K}\mathrm{Kg}{\mathrm{mol}}^{-1}$

Given,

Depression in the freezing point of water = $2^{\mathrm{o}}\mathrm{C}$

Substituting in the equation we have,

$2=2\times 1.86\times \mathrm{m}$

$\mathrm{m}=0.538\mathrm{mol}{\mathrm{Kg}}^{-1}$

Step 3: Calculating the freezing point

Equimolar implies that molality of $\mathrm{NaCl}$ = molality of ${\mathrm{MgCl}}_2$

Since for${\mathrm{MgCl}}_2$, i=3

Substituting the value in equation (1),

we get,

Depression in freezing point= $3\times 1.86\times 0.538=3.002$

So, the freezing point of ${\mathrm{MgCl}}_2$ solution = $-3.{002}^{o}C$