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Logical ReasoningLevel 2 · Application

A liquid boils when its vapour pressure equals atmospheric pressure. A dissolved non-volatile solute always lowers the vapour pressure of the solvent. Without any equations, predict: will a solution have a higher or lower boiling point than the pure solvent, and why?

✦Real Life Hook

Car engines run at temperatures above 100°C. If the coolant were pure water, it would boil away in minutes. Engine coolant (ethylene glycol + water) has a boiling point elevated to ~108°C and a freezing point depressed to −37°C. A single solution — one compound — provides protection against both summer boiling and winter freezing. This is boiling point elevation and freezing point depression working simultaneously, designed precisely using $\Delta T_b = K_b m$ and $\Delta T_f = K_f m$ .

The Equation

$\Delta T_b = T_b - T_b^\circ = K_b \cdot m$

where:

$\Delta T_b$ = elevation of boiling point (K or °C)
$T_b^\circ$ = boiling point of pure solvent
$T_b$ = boiling point of solution ( $> T_b^\circ$ )
$K_b$ = ebullioscopic constant (molal boiling point elevation constant) — depends only on the solvent
$m$ = molality of solution (mol of solute per kg of solvent)

Molality formula expanded:
$\Delta T_b = K_b \times \frac{w_2 \times 1000}{M_2 \times w_1}$

where $w_2$ = mass of solute (g), $M_2$ = molar mass of solute, $w_1$ = mass of solvent (g).

Ebullioscopic constants (Kb) and normal boiling points of common solvents

Solvent	Boiling Point (°C)	Kb (K·kg/mol)
Water	100.0	0.52
Benzene	80.1	2.53
Ethanol	78.4	1.20
Chloroform	61.2	3.63
Carbon tetrachloride	76.7	5.03

📖NCERT 2.8NCERT Intext

Problem

1.8 g of glucose ( $\ce{C6H12O6}$ , $M = 180$ g/mol) is dissolved in 100 g of water. Find the boiling point of the solution. ( $K_b$ for water = 0.52 K·kg/mol)

📖NCERT 2.9NCERT Intext

Problem

The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute is dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. ( $K_b$ for benzene = 2.53 K·kg/mol)

✦JEE / NEET Exam InsightJEE / NEET

◆ΔTb is always positive — boiling point always rises when solute is added.

◆Unit trap:

K_b

is in K·kg/mol. Molality must be in mol/kg. Mass of solvent must be in kg in the formula (or use the expanded form with 1000 factor).

◆Electrolyte trap: For electrolytes, use effective molality =

i \times m

where

i

is van't Hoff factor. NaCl:

\Delta T_b = 2 \times K_b \times m

◆Molar mass shortcut:

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

Quick Check

Q1.What does the ebullioscopic constant (Kb) of a solvent physically represent?

Logical ReasoningLevel 2 · Application

✦Real Life Hook

The Equation

$\Delta T_b = T_b - T_b^\circ = K_b \cdot m$

where:

$\Delta T_b$ = elevation of boiling point (K or °C)
$T_b^\circ$ = boiling point of pure solvent
$T_b$ = boiling point of solution ( $> T_b^\circ$ )
$K_b$ = ebullioscopic constant (molal boiling point elevation constant) — depends only on the solvent
$m$ = molality of solution (mol of solute per kg of solvent)

Molality formula expanded:
$\Delta T_b = K_b \times \frac{w_2 \times 1000}{M_2 \times w_1}$

where $w_2$ = mass of solute (g), $M_2$ = molar mass of solute, $w_1$ = mass of solvent (g).

Ebullioscopic constants (Kb) and normal boiling points of common solvents

Solvent	Boiling Point (°C)	Kb (K·kg/mol)
Water	100.0	0.52
Benzene	80.1	2.53
Ethanol	78.4	1.20
Chloroform	61.2	3.63
Carbon tetrachloride	76.7	5.03

📖NCERT 2.8NCERT Intext

Problem

1.8 g of glucose ( $\ce{C6H12O6}$ , $M = 180$ g/mol) is dissolved in 100 g of water. Find the boiling point of the solution. ( $K_b$ for water = 0.52 K·kg/mol)

📖NCERT 2.9NCERT Intext

Problem

Quick Check

Q1.What does the ebullioscopic constant (Kb) of a solvent physically represent?