Raoult’s Law Examples

Some examples of Raoult’s Law are,

Example 1: 2 moles of P are mixed with 4 moles of Q to form a liquid solution, determine the total vapour pressure of the solution given that vapour pressures of P and Q in pure form are 70 torr and 40 torr respectively.

Solution:

Here, mole fraction of P, x_p = 2/(2+4) = 1/3 and mole fraction of Q, x_q = 4/(2+4) = 2/3

According to Raoult’s law,

p_Total = p₁⁰x₁ + p₂⁰x₂

p_Total = 70×(1/3) + 40×(2/3)

p_Total = 50

Thus, total vapour pressure of the solution is 50 torr as determined by Raoult’s law.

Example 2: Determine the vapor pressure of a solution containing 600 g of n-heptane and 288 g of n-pentane at 25^oC. The vapor pressure of n-heptane and n-pentane are 46 mm Hg and 512 mm Hg respectively.

Solution:

We know that,

Molecular weight of n-heptane = 100 g

No. of Moles of n-heptane in the solution = 600/100 = 6

Molecular weight of n-pentane = 72

No. of moles of n-pentane in the solution = 288/72 = 4

Now,

Mole Fraction of n-heptane = 6/(6+4) = 6/10 = 0.6

Mole Fraction of n-pentane = 4/(6+4) = 4/10 = 0.4

As solution of n-pentane and n-heptane forms a nearly ideal solution, using Raoult’s Law here,

According to Raoult’s law,

p_Total = p₁⁰x₁ + p₂⁰x₂

p_Total = 0.6×46 + 0.4×1400

p_Total = 587.6 mm Hg

Thus, total vapour pressure of the solution is obtained as 587.6 mm Hg using Raoult’s law.

Raoult’s Law in chemistry relates partial pressures of volatile liquid components to their mole fractions in a liquid solution. It states that the partial pressure of each component in the solution is directly proportional to its mole fraction. Thus, it helps us to calculate the total vapour pressure of the solution. Based on Raoult’s law, liquid solutions are classified as Ideal Solutions and Non-Ideal Solutions.

In this article, we will discuss the definition of Raoult’s law, ideal and non-ideal solutions, Raoult’s law for non-volatile solutes and some solved numerical problems based on Raoult’s law.