Animated Solution for Chemistry - Solutions: When $3.00\text{ g}$ of a substance 'X' is dissolved in $100\text{ g}$ of ${\text{CCl}}_4$, it raises the boiling point by $0.60\text{ K}$. The molar mass of the substance 'X' is ......... ${\text{g mol}}^{-1}$. (Nearest integer) [Given, $K_b$ for ${\text{CCl}}_4$ is $5.0{\text{ K kg mol}}^{-1}$]

The phenomenon of boiling point elevation is one of the most fascinating and practically useful colligative properties in chemistry. When you add a non-volatile solute to a pure solvent, the vapor pressure of the solvent decreases. As a result, the solution must be heated to a higher temperature to make its vapor pressure equal to the atmospheric pressure. This increase in the boiling temperature is what we call the elevation in boiling point.

In this problem, we are given a classic laboratory scenario. We have a beaker containing $100\text{ g}$ of carbon tetrachloride (${\text{CCl}}_4$) acting as our solvent. To this, we add $3.00\text{ g}$ of an unknown substance 'X'. We observe that the boiling point of the solution rises by $0.60\text{ K}$. Our mission is to determine the molar mass of this mysterious substance 'X'.

To connect the macroscopic observation (the rise in boiling point) to the microscopic property (the molar mass of the solute), we rely on the mathematical expression for boiling point elevation:

Here, $\Delta T_b$ is the elevation in boiling point, $K_b$ is the ebullioscopic constant (or molal elevation constant) of the solvent, and $m$ is the molality of the solution.

Molality is a measure of concentration defined as the number of moles of solute per kilogram of the solvent. It is particularly useful in thermodynamics because, unlike molarity, it does not change with temperature. We can expand the molality term as follows:

Where $w$ is the given mass of the solute, $M$ is its molar mass, and $W_{\text{solvent}}$ is the mass of the solvent in kilograms.

Now, let's carefully substitute the known values into our expanded equation. We are given: - $\Delta T_b=0.60\text{ K}$ - $K_b=5.0{\text{ K kg mol}}^{-1}$ - Mass of solute ($w$) = $3.00\text{ g}$ - Mass of solvent = $100\text{ g}=0.100\text{ kg}$

Plugging these into the formula, we get:

Let's simplify the right side of the equation. Multiplying the numerator terms, $5.0\times 3.00$ gives $15.0$. The equation becomes:

To isolate $M$, we cross-multiply:

Finally, dividing both sides by $0.060$:

Thus, the molar mass of the unknown substance 'X' is $250{\text{ g mol}}^{-1}$.

This problem beautifully demonstrates how a simple temperature measurement can reveal the fundamental properties of a molecule. Always remember to convert the mass of the solvent into kilograms when calculating molality to avoid silly mistakes!