Understanding the Relationship Between Molar Mass and Ebullioscopic Constant

The question asks us to identify the correct relationship between the molar mass of a solute and the Ebullioscopic constant. This relationship comes from the study of colligative properties, specifically the elevation of the boiling point of a solvent when a non-volatile solute is dissolved in it.

What is Elevation in Boiling Point?

When a non-volatile solute is added to a pure solvent, the boiling point of the solution increases compared to the pure solvent. This increase in boiling point is called the elevation in boiling point, denoted by $\Delta T_b$. It is a colligative property, meaning it depends on the number of solute particles in the solution, not their identity.

Relating Elevation in Boiling Point to Molality

The elevation in boiling point ($\Delta T_b$) is directly proportional to the molality ($m$) of the solution for dilute solutions.

Mathematically, this is expressed as:

$\Delta T_b \propto m$

To remove the proportionality and introduce an equality, we use a constant known as the Ebullioscopic constant, or molal boiling point elevation constant, denoted by $K_b$.

So, the relation becomes:

\( \Delta T_b = K_b \times m \)

Defining Molality (m)

Molality is defined as the number of moles of solute per kilogram of solvent.

\( m = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} \)

Let's denote:

• Mass of solute = \(w_2\) (in grams)
• Molar mass of solute = \(M_2\) (in g/mol)
• Mass of solvent = \(w_1\) (in grams)

The number of moles of solute is given by:

\( \text{Moles of solute} = \frac{w_2}{M_2} \)

The mass of solvent in kilograms is given by:

\( \text{Mass of solvent in kg} = \frac{w_1}{1000} \)

Now, substitute these into the definition of molality:

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

Deriving the Formula for Molar Mass (\(M_2\))

We have the relationship:

\( \Delta T_b = K_b \times m \)

Substitute the expression for molality ($m$) into this equation:

\( \Delta T_b = K_b \times \frac{1000 \times w_2}{M_2 \times w_1} \)

Now, we need to rearrange this equation to solve for the molar mass of the solute, \(M_2\). We can do this by cross-multiplication:

\( \Delta T_b \times (M_2 \times w_1) = K_b \times (1000 \times w_2) \)

\( M_2 \times \Delta T_b \times w_1 = 1000 \times w_2 \times K_b \)

Finally, isolate \(M_2\) by dividing both sides by \( \Delta T_b \times w_1 \):

\( M_2 = \frac{1000 \times w_2 \times K_b}{\Delta T_b \times w_1} \)

Comparing with the Options

Let's compare our derived formula with the given options:

Our derived formula matches option 3.

Revision Table: Key Formulas for Boiling Point Elevation

Additional Information: Ebullioscopic Constant and Colligative Properties

The Ebullioscopic constant ($K_b$) is a property of the solvent, not the solute. Each solvent has a specific $K_b$ value, which represents the boiling point elevation for a 1 molal solution of a non-volatile solute in that solvent. For water, the value of $K_b$ is approximately 0.512 °C kg/mol.

Boiling point elevation is one of four main colligative properties. The others are:

• Relative Lowering of Vapor Pressure: Lowering of vapor pressure of a solvent when a non-volatile solute is added.
• Depression in Freezing Point: Lowering of the freezing point of a solvent when a non-volatile solute is added (related to the Cryoscopic constant, \(K_f\)).
• Osmotic Pressure: The pressure required to prevent the flow of solvent across a semipermeable membrane into a solution.

These properties are useful for determining the molar mass of unknown solutes, especially macromolecules like polymers, as long as the solute is non-volatile and does not dissociate or associate in the solvent (or if its Van't Hoff factor is known).