Question

Derive the expression for molar mass of solute in terms of boiling point elevation of solvent.

Show how the molecular mass of a solute can be determined on the basis of elevation of boiling point of the solution.

Derive an expression for determining the molecular mass of a solute on the basis of elevation of boiling point.

Explain the relevance of elevation of boiling point in determining the molecular weight of a non-volatile solute.

Answer 1

The boiling point elevation is directly proportional to the molality of the solution. Thus,

∆Τ_b = K_b m    ...(1)

Suppose we prepare a solution by dissolving W₂ g of solute in W₁ g of solvent. Moles of solute in W₁ g of solvent = `W_2/M_2`

where, M₂ is the molar mass of solute.

Mass of solvent = W₁ g

= `(W_1  "g")/(1000  "g"//"kg")`

= `W_1/1000` kg

The molality is expressed as,

m = `"Moles of solute"/"Mass of solvent in kg"`

m = `(W_2//M_2  "mol")/(W_1//1000  "kg")`

m = `(1000 W_2)/(M_2 W_1)` mol kg⁻¹    ...(2)

Substituting equation (2) in equation (1), we get,

`Delta T_b = (1000 K_b W_2)/(M_2 W_1)`

Hence,

`M_2 = (1000 K_b W_2)/(Delta T_b W_1)`

Answer 2

The molecular mass of a solute dissolved in a particular solvent can be determined with the help of the elevation of the boiling point for the solution as follows.

We know that

ΔT_b = K_b . m    ...(i)

If w g of a solute are dissolved in W g of a solvent, the molality m of the solution is given by

`m = "Number of moles of solute"/"Mass of solvent in kg"`

= `(w//M')/(W//1000)`

or `m = (1000 xx w)/(W xx M0')`

where M' is the molecular mass of the solute.

Putting the value of m in equation (i), we have

`Delta T_b = K_b xx (1000 xx w)/(W xx M')`

or `M' = (1000 xx K_b xx w')/(W xx M')`