Ideal Solution - Meaning, Definition, Examples, FAQs

Ideal Solution

An ideal solution is a solution that obeys Raoult's law over the entire range of composition. The behavior of such solutions closely matches the theoretical predictions made by Raoult's law.

Characteristics of Ideal Solutions

1. Similar Intermolecular Interactions

In an ideal solution, the intermolecular forces between unlike molecules are nearly equal to those between like molecules.

A–B ≈ A–A ≈ B–B

where:

• A–A = interactions between solvent molecules
• B–B = interactions between solute molecules
• A–B = interactions between solute and solvent molecules

Since the newly formed A–B interactions are almost identical to the original interactions, mixing occurs without any significant energy change.

2. Enthalpy of Mixing

No heat is absorbed or evolved during the mixing process.

$\Delta H_{\operatorname{mix}}=0$.

This means the enthalpy of the solution remains unchanged after mixing.

3. Volume of Mixing

The total volume of the solution is equal to the sum of the volumes of the individual components.

$\Delta V_{\operatorname{mix}}=0$

For example, if 1 litre of liquid A is mixed with 1 litre of liquid B to form an ideal solution, the final volume of the solution will be exactly 2 litres.

4. Entropy of Mixing

Mixing increases the randomness of the system because molecules become more uniformly distributed throughout the solution.

$\Delta S_{\operatorname{mix}}>0$

Therefore, the entropy of mixing is always positive.

5. Gibbs Free Energy of Mixing

Since the entropy of mixing is positive and the enthalpy change is zero, the mixing process occurs spontaneously.

$\Delta G_{\operatorname{mix}}<0$

Thus, the formation of an ideal solution is a spontaneous process.

Related Topics Link

• Mixtures
• Solute
• Solution Properties Concentration
• Relation Between Molarity And Molality
• Osmotic Pressure Equation

Vapour Pressure of Ideal Solutions

Ideal solutions obey Raoult's law, according to which the partial vapour pressure of each component is directly proportional to its mole fraction in the solution.

$\begin{aligned}
& P_A=P_A^0 X_A \\
& P_B=P_B^0 X_B
\end{aligned}$

where:

• $P_A$ and $P_B$ are the partial vapour pressures of components A and B ,
• $P_A^0$ and $P_B^0$ are the vapour pressures of pure A and pure B ,
• $X_A$ and $X_B$ are their mole fractions in the solution.

Examples of Ideal Solutions

For the solutions to follow the ideal solution at all ranges of concentrations and temperatures then the molecular size of the liquids should be nearly the same.
For example:

• CH₃OH + C₂H₅OH: Both these liquids are polar and have nearly the same size. Thus, this solution is an ideal solution.
• C₂H₅Br₂ + C₂H₅Cl₂: Both these liquids are polar and have nearly the same size. Thus, this solution is an ideal solution.
• C₂H₅Cl + C₂H₅Br: Both these liquids are polar and have nearly the same size. Thus, this solution is an ideal solution.
• C₆H₆ + C₆H₅CH₃: Both these liquids are non-polar and have nearly the same size. Thus, this solution is an ideal solution.
• C₂H₅Cl + C₂H₅I: Both these liquids are polar but the size difference between the molecules is large. Thus, this solution is not an ideal solution.

Also Read:

Some Solved Example

Question 1: Equimolal solutions in the same solvent have

1)same boiling point but different freezing point

2)same freezing point but different boiling point

3) (correct)same boiling and same freezing points

4)different boiling and different freezing point

Solution:

Raoult's Law -The total vapour pressure of the binary mixture of miscible liquids ideally is given by

$P_T=P_A^0 x_A+P_B^0 x_B$

Where $x_A$ and $x_B$ are mole fractions of A and B in the liquid phase

$P_{A \text { and }}^0 P_B^0$ are vapour pressures of pure liquids.

According to Raoult's law, equimolar solutions of all substances in the same solvent will show the equal value of colligative properties such as elevation in Boiling Point, depression in freezing point, osmotic pressure and relative lowering of vapour pressure.

Hence, the correct answer is the option (3).

Question 2: A mixture of 100 m mol of $\mathrm{Ca}(\mathrm{OH})_2$ and 2 g of sodium sulphate was dissolved in water and the volume was made up to 100 mL . The mass of calcium sulphate formed and the concentration of $\mathrm{OH}^{-}$ in the resulting solution, respectively , are : (Molar mass of $\mathrm{Ca}(\mathrm{OH})_2, \mathrm{Na}_2 \mathrm{SO}_4$ and $\mathrm{CaSO}_4$ , are 74, 143 and 136 g $\mathrm{mol}^{-1}$ , respectively ; $K_{\text {sp }}$ of $\mathrm{Ca}(\mathrm{OH})_2$ is $5.5 \times 10^{-6}$,)

1) (correct)$1.9 \mathrm{~g}, 0.28 \mathrm{~mol} \mathrm{~L}^{-1}$

2)$13.6 \mathrm{~g}, \quad 0.28 \mathrm{molL}^{-1}$

3)$1.9 \mathrm{~g}, 0.14 \mathrm{molL}^{-1}$

4)$13.6 \mathrm{~g}, \quad 0.14 \mathrm{molL}^{-1}$

Solution:

Given,

Mol of Na₂SO₄ = 2/142 = 14 m mol

$\begin{aligned}
& \mathrm{Ca}(\mathrm{OH})_2+\mathrm{Na}_2 \mathrm{SO}_4 \longrightarrow \mathrm{CaSO}_4+2 \mathrm{NaOH} \\
& \begin{array}{llll}
\mathrm{mmol} & 100 \quad 14 & 14 \mathrm{~m} / \mathrm{mol} \quad 28 \mathrm{~m} / \mathrm{mol}
\end{array} \\
&
\end{aligned}$

Mass of $\mathrm{CaSO}_4=\frac{14 \times 136}{1000}=1.9 \mathrm{gm}$
Molarity of $\mathrm{OH}^{-}=\frac{28}{100}=0.28 \mathrm{~mol} / \mathrm{L}$

Question 3: All of the following form ideal solutions except:

1)$\mathrm{C}_2 \mathrm{H}_5 \mathrm{Br}$ and $\mathrm{C}_2 \mathrm{H}_5 \mathrm{I}$

2)$\mathrm{C}_6 \mathrm{H}_5 \mathrm{Cl}$ and $\mathrm{C}_6 \mathrm{H}_5 \mathrm{Br}$

3)$\mathrm{C}_6 \mathrm{H}_6$ and $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_3$

4) (correct)$\mathrm{C}_2 \mathrm{H}_5 \mathrm{I}$ and $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$

Solution:

$\mathrm{C}_2 \mathrm{H}_5 \mathrm{I}$ and $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$ do not form an ideal solution.

Hence, the correct answer is the option (4).

Question 3: Which of the following is incorrect?

1)Relative lowering of vapour pressure is independent of the nature of the solute

2)The vapour pressure is not a colligative property.

3) The vapour pressure of a solution is lower than the vapour pressure of the solvent

4) (correct)The relative lowering of vapour pressure is directly proportional to the original pressure

Solution:

The vapour pressure is not a colligative property but lowering of Vapor Pressure is a colligative property of solutions.

According to Raoult's law, the relative lowering in vapour pressure of a dilute solution is equal to the mole fraction of the solute present in the
solution.

Hence, the correct answer is the option (4).

Question 5: 100 mL of liquid A and 25 mL of liquid B are mixed to form a solution of volume 125 mL. Then the solution is:

1) (correct)Ideal

2) Non-ideal with positive deviation

3) Non-ideal with negative deviation

4)Cannot be predicted

Solution:

Here,
V_A = 25 mL , V_B = 100 mL
After mixing
V_A + V_B = 125 mL
then,
$\Delta V_{\operatorname{mix}}=125-(100+25)=0$
Hence, the solution is ideal.
Hence, the correct answer is the option (1).

Question 6: Liquid and liquid $' N^{\prime}$ form an ideal solution. The vapour pressures of pure liquids $' M^{\prime}$ and $' N^{\prime}$ are $450$ and 700 mmHg, respectively, at the same temperature. Then correct statement is :

$x_M=$ Mole fraction of ${ }^{\prime} M^{\prime}$ in solution;

$x_N=$ Mole fraction of ${ }^{\prime} N^{\prime}$ in solution;

$y_M=$ Mole fraction of $' M$ ' in vapour phase;

$y_N=$ Mole fraction of ${ }^{\prime} N^{\prime}$ in vapour phase)

1) $\frac{x_M}{x_N}=\frac{y_M}{y_N}$

2)$\left(x_M-y_M\right)<\left(x_N-y_N\right)$

3) $\frac{x_M}{x_N}<\frac{y_M}{y_N}$

4) (correct) $\frac{x_M}{x_N}>\frac{y_M}{y_N}$

Solution:

The vapour pressures of pure liq. M & N are 450 mm of Hg and 700 mm of Hg respectively,

$\begin{aligned} & P_N^0>P_M^0 \\ & y_N>x_N\end{aligned}$ $(N \rightarrow$ more volatile)

$\begin{aligned} & y_M<x_M \Rightarrow x_M>y_M \\ & y_N<x_N \rightarrow x_N>y_N \\ & \frac{x_M}{x_N}>\frac{y_M}{y_N}\end{aligned}$

Hence, the answer is the option (4).

Practice More Question from the link given below: