Question

A solution of 1-propanol and 2-propanol having $\dfrac{3}{4}$ by mass of 2-propanol has an equilibrium vapour pressure of 88.8 mm Hg. Another solution having $\dfrac{1}{3}$ by mass of 2-propanol has an equilibrium vapour pressure of 68.3 mm Hg. Calculate the vapour pressure of pure alcohols at ${40^ \circ }C$ assuming ideal solution mixture prepared at ${40^ \circ }C$.

Answer 1

According to Dalton’s law,
${p_{total}} = {p_1} + {p_2}$
where ${p_1}{\text{ and }}{{\text{p}}_2}$ are the partial pressures of the components. The Raoult’s law states that the partial pressures of the components are proportional to their mole fraction.

Complete step by step answer:
We are given the vapour pressures of the solution of 1-propanol and 2-propanol. We will find the vapour pressure of 1-propanol and 2-propanol by using Raoult's law.
- We know that 1-propanol and 2-propanol are isomers and their molecular formula is ${C_3}{H_8}O$. So, their molecular mass will be the same.
Now, according to Dalton’s law of partial pressures, it states that the total pressure of the solution
${p_{total}} = {p_1} + {p_2}{\text{ }}...{\text{(1)}}$
Here, ${p_1}$ is the partial pressure of 1-propanol and ${p_2}$ is the partial pressure of 2-propanol.
According to Raoult’s law the partial pressure of a component
$p = X \cdot {p^ \circ }{\text{ }}....{\text{(2)}}$
Where X is the mole fraction is the component and ${p^ \circ }$ is the vapour pressure of the component.
Putting the equation (2) into equation (1), we get
${p_{total}} = {X_1} \cdot p_1^ \circ + {X_2} \cdot p_2^ \circ {\text{ }}...{\text{(3)}}$
Now, in one case we are given that the mass proportion of 2-propanol is $\dfrac{3}{4}$ in its mixture with 1-propanol and the total vapour pressure is 88.8 mm Hg. Here, as the molecular masses of both of them are the same, the mole fractions of them will be equal to its mass proportions.
So, we can say that ${X_2}$ will be $\dfrac{3}{4}$ and ${X_1}$ will be $1 - \dfrac{3}{4} = \dfrac{1}{4}$ . So, putting all the available values into equation (3), we get
$\Rightarrow 88.8 = \left( {\dfrac{1}{4}} \right)p_1^ \circ + \left( {\dfrac{3}{4}} \right)p_2^ \circ {\text{ }}.......{\text{(4)}}$
Now, in another case, we are given that the proportions are $\dfrac{1}{3}$ by mass of 2-propanol. So, mass proportions of 1-propanol will be $1 - \dfrac{1}{3} = \dfrac{2}{3}$ and the total vapour pressure of the solution is 68.3 mm Hg. So, we can put all the available values into equation (3) as
$\Rightarrow 68.3 = \left( {\dfrac{2}{3}} \right)p_1^ \circ + \left( {\dfrac{1}{3}} \right)p_2^ \circ {\text{ }}.......{\text{(5)}}$
Now, as we solve the equations (4) and (5), we get that $p_1^ \circ$ is 52 mm Hg and $p_2^ \circ$ is 75 mm Hg.
Thus, we obtained the vapour pressure of 1-propanol = 52 mm Hg and vapour pressure of 2-propanol = 75 mm Hg.

Note: Here, we are given the vapour pressure in mmHg unit and we do not need to convert it into the SI unit of pressure. Note that as the molar masses of the two compounds are same, then their mass ratios will be same as their mole fractions in their mixture.