Elevation in Boiling Point

Have you observed that if you add salt to boiling water, it will boil slightly faster? This simple observation is a demonstration of one of the basics of solution chemistry known as Elevation in Boiling Point. Let's explore this scientifically valuable and useful concept of solutions in detail.

1.0Introduction to Boiling Point: Basic Concept 

The boiling point of a liquid is the temperature at which its vapour pressure is the same as atmospheric pressure so that it can change into vapour at all points of its body. In simple words, it is the temperature at which the droplets of water start to convert into vapour due to the application of heat. Every liquid has a certain boiling point at standard atmospheric pressure; for example, pure water boils at 100°C (373 K) under ordinary atmospheric pressure.

2.0Elevation in Boiling Point Definition: 

Elevation in boiling point is the increase in the boiling point of a certain solvent when a non-volatile solute is added to it. For example, in the case of salt and water, water is the solvent, and salt is the non-volatile solute. This effect is a result of decreased vapour pressure caused by the presence of added or extra solute particles. Hence, the elevation in boiling point definition can be stated as: 

“It is the phenomenon in which the boiling point of a solvent rises when a non-volatile solute is added to it to produce a solution.” 

3.0Elevation in the Boiling Point Formula: 

The elevation in boiling point formula is useful for highlighting the relationship between temperature change and concentration of a solvent. The formula for elevation in boiling point can be expressed as: 

$\Delta T_b=K_b\cdot m$

Here, 

• ΔT_b​ = Elevation in boiling point (in K or °C)
• K_b = Molal boiling point elevation constant (depends on solvent), also known as ebullioscopic constant. 
• m = Molality of the solution (mol/kg)

4.0Boiling Point Elevation Graph: 

A standard boiling point elevation graph is the visual representation of the boiling point temperature (x-axis) and vapour pressure (y-axis) of a specific solvent. The graph is given as: 

This diagram illustrates that the vapour pressure curve of the solution is below that of the pure solvent at any temperature because the presence of solute particles lowers the escaping tendency of the solvent molecules. Consequently, the solution needs a greater temperature to achieve atmospheric pressure, so its boiling point is moved to the right. The horizontal distance between the two boiling points on the graph is the elevation in boiling point (ΔT_b), which shows how the solution boils at a greater temperature than the pure solvent.

5.0Factors Affecting Elevation in Boiling Point: 

The rate of boiling point does not increase or decrease on its own; there are a number of factors that affect the elevation of the boiling point of a solution. Some of these factors affecting elevation in boiling point are:

1. Nature of the solute: Ionic solutes, such as NaCl, break down into more particles, resulting in a higher elevation.
2. Concentration of the solute: Increased concentration of solute results in an increased molality, leading to an increased elevation in the boiling point of the solvent.
3. Type of solvent: Solvents with larger boiling point elevation constants (Kb) exhibit more effect.
4. Ionisation/association of solute: Strong electrolytes (such as HCl) ionise totally and increase particle number. Weak electrolytes or associating solutes (such as acetic acid, benzoic acid in benzene) will partially ionise or dimerise, decreasing the effective particle number and consequently causing a boiling point elevation.
5. Atmospheric Pressure: Although not directly, atmospheric pressure determines the baseline of the boiling point, which ultimately helps in measuring the elevation of the boiling point. 

6.0Colligative Properties and Elevation in Boiling Point: 

Colligative properties are the physical properties of solutions that depend on the number of particles of a solute. Boiling point elevation possesses the following colligative properties of a solvent: 

1. Relative Reduction of Vapour Pressure: Solute particles lower the surface area for solvent molecules to come out, reducing vapour pressure.
2. Increase in Boiling Point: Because the vapour pressure is reduced, the solution is required to be heated more to achieve atmospheric pressure, so the boiling point increases.
3. Decrease in Freezing Point: Solute particles hinder the creation of the solid phase, and hence, the freezing point decreases.
4. Osmotic Pressure: Pressure required to prevent solvent flow through a semipermeable membrane into the solution; rises with additional solute particles.

7.0Example of Elevation in Boiling Point: 

Problem 1: Calculate the elevation in boiling point when 18 g of glucose (C₆H₁₂O₆) is dissolved in 500 g of water. (Given: Molar mass of glucose = 180 g/mol, K_b for water = 0.512 K·kg/mol). 

Solution: According to the question

Moles of glucose ​​$=\frac{18}{180}=0.1\mathrm{mol}$

Mass of solvent in Kg $=\frac{500}{1000}=0.5\mathrm{kg}$

Molality (m) $=\frac{0.1}{0.5}=0.2\mathrm{mol}/\mathrm{kg}$

Now, using the elevation in boiling point formula: 

$\begin{array}{rl}& \Delta T_b=K_b\cdot m\\ & \Delta T_b=0.512\cdot 0.2=0.1024K\end{array}$

Hence, the elevated boiling of the solution = 100 + 0.1024 = 100.10 °C. 

Problem 2: Calculate the boiling point of a 4% by weight urea solution in water. Take K_b for water = 0.512 K·kg/mol and the boiling point of water 100 °C. 

Solution: Let the total mass of the solution = 100g 

Hence, the mass of urea (solute) = 4g 

Mass of water (Solvent) = 100 – 4 = 96 g = 0.096 kg

Molar mass of urea (NH₂CONH₂) = 60 g/mol

Hence, Moles of $urea=\frac{4}{60}=0.0667\mathrm{mol}$

Molality of the solution $(m)=\frac{0.0667}{0.096}\approx 0.694\mathrm{mol}/\mathrm{kg}$

Now, applying the formula for the elevation of the boiling point: 

$\begin{array}{rl}\Delta T_b& =K_b\cdot m\\ \Delta T_b& =0.512\cdot 0.694=0.355K\end{array}$

Hence, the final boiling point of water with urea = 100 + 0.355 = 100.36 °C. 

Problem 3: 10 g of a non-volatile, non-dissociating solute is dissolved in 250 g of toluene. The solution boils at 111.1°C. The boiling point of pure toluene is 110.6°C, and the molal boiling point elevation constant (K_b) for toluene is 0.60 K·kg/mol. Calculate the molar mass of the solute.

Solution: Here, the elevation in the boiling point of the solution: 

$\Delta T_b=111.1-110.6=0.5^{\circ }\mathrm{C}$

Hence, the molality of the solution: 

$\begin{array}{rl}& \Delta T_b=K_b\cdot m\\ & 0.5=0.60\cdot m\\ & m=\frac{0.5}{0.60}=0.833\mathrm{mol}/\mathrm{kg}\end{array}$

Given that the mass of solvent (toluene) = 250 g = 0.25 kg 

We know, 

$\text{ Molality }(\mathrm{m})=\frac{\text{ moles of solute }}{\mathrm{Kg}\text{ of solvent }}$

Hence, 

$\begin{array}{rl}& \text{ Moles of Solute }=\mathrm{m}\times \mathrm{kg}\text{ of solute }\\ & \text{ Moles of Solute }=0.833\times 0.25=0.20825\mathrm{mol}\end{array}$

Now, calculate the molar mass of the solute: 

$\text{ Molar mass }=\frac{\text{ Mass of solute }}{\text{ moles of solute }}=\frac{10}{0.20825}\approx 48.01\mathrm{g}/\mathrm{mol}$