Determination of Molecular Weight using Colligative Properties

Let us consider the colligative properties one by one and molecular weight calculation based on these properties.

1. Elevation of Boiling Point

\( ∆T_b=K_b m \)

Where,

\( ΔT_b \) = elevation in boiling point
\(K_b\) = Boiling Point Elevation Constant

m = molal concentration of the solutionNow we know that molality is the number of moles of solute per

Now we know that molality is the number of moles of solute per kilogram of the solvent. Let the weight of solute be w2 having molar mass M2 dissolved in w1 grams of solvent. Then molality can be given as:

Then molality can be given as: \(m= \frac {moles ~of~ solute}{mass ~of~ solvent~ in ~Kg} = \frac {w_2/M_2}{w_1/1000} \)

\(m= \frac {moles ~of~ solute}{mass ~of~ solvent~ in ~Kg} = \frac {w_2/M_2}{w_1/1000} \)

\( \Rightarrow  m = \frac {1000 × w_2}{w_1 × M_2} \)So now, elevation of boiling point is given by:

So now, elevation of boiling point is given by:

Now if we rearrange the above equation, we get:

The above equation gives us a relation between molecular weight and the colligative property that is elevation in boiling point.

2.   Depression of Freezing Point

The depression in freezing point is another colligative property of solutions that gives us a method to determine the molar masses of various substances similar to elevation of boiling point. We know that depression in freezing point is given as:

\( ∆T_b=K_b m \)

Where,

From the above description we now know that molality is given as:

\( m = \frac {1000 × w_2}{w_1 × M_2} \)

\( ∆T_f = \frac { K_f × 1000 × w_2}{w_1 × M_2 } \)

Now if we rearrange the above equation, we get:

\( => M_2 = \frac {K_f × 1000 × w_2}{w_1 × ∆T_f} \)

3.Osmotic Pressure

Mathematically it is given as:

π=CRT

\( C = \frac {w_2/M_2}{V} = \frac {w_2}{V × M_2 } \)So we can now write the osmotic pressure as:

So we can now write the osmotic pressure as:

\( π = \frac {w_2 RT}{M_2 V} \)

Rearranging the above equation we get,

\( M_2 = \frac {w_2 RT}{πV} \)