When 1.5 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point of benzene got raised from 353.23 K to 353.93 K. Calculate the molar mass of the solute. (Given $K_b$ for benzene = 2.52 K kg mol$^{-1}$)

When 1.5 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point of benzene got raised

Question

K_b

CBSECBSE Class XII Board 2026Subjective· 3mImportance★★★★★

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Concept understanding — Boiling Point Elevation

Boiling Point Elevation — From Intuition to Precision

Imagine you're cooking pasta. You add salt to the water, and the water takes longer to boil. That's not your imagination — it's a real physical effect. The salt has raised the boiling point of the water. Pure water boils at 100°C at sea level, but salt water needs a slightly higher temperature to boil. That's boiling point elevation in action.

Why does this happen?

To understand why, you need to think about what boiling actually is. Boiling occurs when the vapour pressure of the liquid equals the surrounding atmospheric pressure. Vapour pressure is the pressure exerted by molecules escaping from the liquid surface into the gas phase.

When you dissolve a non-volatile solute (like salt, which doesn't evaporate) in a solvent (like water), something interesting happens at the surface. Some of the surface molecules are now solute particles — they don't escape into the vapour. This means fewer solvent molecules can leave the liquid per second. The result? The vapour pressure of the solution is lower than that of the pure solvent at the same temperature.

Now, to make this reduced vapour pressure equal to the atmospheric pressure (so the liquid can boil), you need to raise the temperature. That extra heat gives the remaining solvent molecules enough energy to overcome the solute's interference and produce the required vapour pressure.

Note

The solute itself doesn't boil away — it stays behind. That's why we call it a non-volatile solute. If the solute were volatile (like alcohol), the story changes.

The precise statement

Boiling point elevation is the increase in the boiling point of a solvent when a non-volatile solute is dissolved in it. The boiling point of the solution is always higher than that of the pure solvent.

The elevation $\Delta T_b$ is given by:

\Delta T_b = T_b(\text{solution}) - T_b(\text{pure solvent})

And it depends only on the number of solute particles, not on their chemical identity (for dilute solutions). This is a colligative property — a property that depends on the concentration of particles, not their nature.

\Delta T_b = K_b \cdot m

What is $K_b$ ?

The ebullioscopic constant $K_b$ is a property of the pure solvent. It tells you how much the boiling point rises per unit molal concentration. For water, $K_b = 0.512 \, \text{°C kg mol}^{-1}$ . So a 1 molal aqueous solution (1 mole of solute per kg of water) boils at $100.512\,°\text{C}$ .

Tip

For quick calculations: $\Delta T_b \propto m$ . Double the molality, double the elevation. But this works only for dilute solutions — at high concentrations, interactions between solute particles break the linearity.

What about electrolytes?

If the solute dissociates into ions (like NaCl → Na⁺ + Cl⁻), the number of particles increases. One mole of NaCl gives two moles of particles in solution. So the effective concentration is higher. We account for this using the van't Hoff factor $i$ :

\Delta T_b = i \cdot K_b \cdot m

For NaCl, $i \approx 2$ (in dilute solutions). For sugar (which doesn't dissociate), $i = 1$ . …

Why this formula?

Boiling Point Elevation: Why the Formula Holds

Let's build this from first principles — understanding the why before the what.

What Happens at the Boiling Point?

At the boiling point, a liquid's vapor pressure equals the external atmospheric pressure. For a pure solvent, this happens at a fixed temperature (e.g., 100°C for water at 1 atm).

When you add a non-volatile solute (like salt or sugar), the solute particles occupy space at the liquid surface, reducing the number of solvent molecules that can escape into vapor. This lowers the vapor pressure of the solution compared to the pure solvent.

The Key Consequence

Since vapor pressure is now lower, the solution must be heated to a higher temperature to make its vapor pressure reach atmospheric pressure again. That's the boiling point elevation.

The Formula and Its Derivation

The boiling point elevation $\Delta T_b$ is given by:

\Delta T_b = K_b \cdot m

where:

$\Delta T_b = T_b(\text{solution}) - T_b(\text{pure solvent})$
$K_b$ = ebullioscopic constant (depends only on the solvent)
$m$ = molality of the solution (moles of solute per kg of solvent)

Why Molality and Not Molarity?

Molality is temperature-independent — it doesn't change when the solution is heated. Molarity (moles per liter) changes because volume expands with temperature. Since we're measuring a temperature change, we need a concentration unit that stays fixed.

The Reasoning Behind $K_b$

The constant $K_b$ comes from thermodynamics. For a dilute solution, the vapor pressure lowering follows Raoult's Law:

P_{\text{solution}} = x_{\text{solvent}} \cdot P_{\text{solvent}}^0

where $x_{\text{solvent}}$ is the mole fraction of solvent and $P_{\text{solvent}}^0$ is the pure solvent's vapor pressure.

Using the Clausius-Clapeyron equation (which relates vapor pressure to temperature) and Raoult's Law, one can derive:

K_b = \frac{R T_b^2}{1000 \cdot \Delta H_{\text{vap}}}

where:

$R$ = gas constant (8.314 J/mol·K)
$T_b$ = boiling point of pure solvent (in Kelvin)
$\Delta H_{\text{vap}}$ = molar enthalpy of vaporization (J/mol)
The factor 1000 converts grams to kg (since molality uses kg of solvent)

What This Tells Us

$K_b$ is a property of the solvent alone — it doesn't depend on the solute.
Solvents with higher boiling points or lower heats of vaporization have larger $K_b$ values (greater elevation per molal concentration).

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The addition of a non-volatile solute lowers the vapor pressure of the solvent, thereby increasing its boiling point. By calculating this boiling point elevation and using the given ebullioscopic constant, we determine the molar mass of the solute to be 60 g mol $^{-1}$ .

When a non-volatile solute is dissolved in a solvent, the vapor pressure of the resulting solution is lower than that of the pure solvent at the same temperature. This is because the solute particles occupy some space at the liquid surface, reducing the number of solvent molecules that can escape into the vapor phase.

Boiling occurs when the vapor pressure of the liquid equals the external atmospheric pressure. Since the solution's vapor pressure is lower, a higher temperature is required for it to reach the atmospheric pressure and boil. This increase in boiling point is known as boiling point elevation, and it is a colligative property, meaning it depends on the concentration of solute particles, not their identity.

The relationship between boiling point elevation and the concentration of the solute is given by the following formula:

\Delta T_b = K_b \cdot m

Molality ( $m$ ) 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)}}

And the moles of solute can be expressed as:

\text{moles of solute} = \frac{\text{mass of solute}}{\text{molar mass of solute}}

Combining these, we can write the boiling point elevation formula in terms of the mass of solute ( $w_2$ ), molar mass of solute ( $M_2$ ), and mass of solvent ( $w_1$ in kg):

\Delta T_b = K_b \cdot \frac{w_2}{M_2 \cdot w_1 \text{ (in kg)}}

We need to calculate $M_2$ .

Here is the step-by-step calculation:

Identify the given values:
- Mass of non-volatile solute ( $w_2$ ) = $1.5 \text{ g}$
- Mass of benzene (solvent, $w_1$ ) = $90 \text{ g}$
- Boiling point of pure benzene ( $T_b^\circ$ ) = $353.23 \text{ K}$
- Boiling point of solution ( $T_b$ ) = $353.93 \text{ K}$
- Molal boiling point elevation constant for benzene ( $K_b$ ) = $2.52 \text{ K kg mol}^{-1}$
Calculate the boiling point elevation ( $\Delta T_b$ ): …

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Q.When 1.5 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point of benzene got raised from 353.23 K to 353.93 K. Calculate the molar mass of the solute. (Given KbK_bKb​ for benzene = 2.52 K kg mol−1^{-1}−1)

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Q.When 1.5 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point of benzene got raised from 353.23 K to 353.93 K. Calculate the molar mass of the solute. (Given $K_b$ for benzene = 2.52 K kg mol $^{-1}$ )

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