Question

In 1000g of liquid B (molar mass 180g/mol), 100g of liquid A (molar mass 140g/mol) was dissolved. The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 torr.

Answer 1

When a liquid is evaporated in a closed vessel, some of the liquid escapes as vapours above the surface of that liquid. Vapour pressure of a liquid can be described as the pressure exerted by these vapours on the liquid in a closed vessel at certain temperature and at equilibrium.

Formula used:

1. ${P_T} = {p_A} + {p_B}$
   The total pressure of the solution containing liquid A and B is denoted by $[{P_T}]$, whereas $[{p_A}]$ and $[{p_B}]$ represents the partial pressure of liquid A and B respectively.
2. ${p_B} = {p^o}B + {X_B}$
   The vapour pressure of liquid B is denoted by $[{p^0}_B]$ and $[{X_B}]$ is mole fraction of liquid B.

Complete step by step answer:
Let us first note down the given quantities in the question.
Mass of liquid A ($[{W_A}]$)=100g
Mass of liquid B ($[{W_B}]$)=1000g
Molar mass of liquid A ($[{M_A}]$)= 140g/mol
Molar mass of liquid A ($[{M_B}]$)= 180g/mol
Total vapour pressure of the solution ($[{P_T}]$)= 475torr
Vapour pressure of liquid B ($[{p^0}_B]$) = 500 torr
To find out the vapour pressure of total solution, we need to know the value of partial pressures of liquid A and B.
Step 1: We first need to find out the number of moles of liquid A and B present in the solution. This will further help us to calculate the mole fraction of liquid A and B in the next step.
We will use the following formula to calculate the number of moles for liquid A and B.
$number{\text{ }}of{\text{ }}moles(n) = \dfrac{{mass}}{{molar{\text{ }}mass}}$
The number of moles of liquid A are:
$number{\text{ }}of{\text{ }}moles{\text{ }}of{\text{ }}liquid{\text{ }}A({n_A}) = \dfrac{{mass{\text{ }}of{\text{ }}liquid{\text{ }}A({W_A})}}{{molar{\text{ }}mass{\text{ }}of{\text{ }}liquid{\text{ }}A({M_A})}} \\\ = \dfrac{{100}}{{140}} = 0.714mol$
Similarly, we can find out the number of moles of liquid B.
$number{\text{ }}of{\text{ }}moles{\text{ }}of{\text{ }}liquid{\text{ }}B({n_B}) = \dfrac{{mass{\text{ }}of{\text{ }}liquid{\text{ }}B({W_B})}}{{molar{\text{ }}mass{\text{ }}of{\text{ }}liquid{\text{ }}B({M_B})}} \\\ = \dfrac{{1000}}{{180}} = 5.556mol$
Step 2: Now, we need to find out the mole fractions of liquid A and B.
${\text{mole fraction of liquid A(}}{{\text{X}}_A}{\text{) = }}\dfrac{{number{\text{ }}of{\text{ }}moles{\text{ }}of{\text{ }}A({n_A})}}{{total\ number\ of\ moles({n_A} + {n_B})}}$
${\text{ = }}\dfrac{{0.714}}{{(0.714 + 5.556)}} = 0.114$
Similarly, we can calculate the mole fraction of liquid B.
${\text{mole fraction of liquid B(}}{{\text{X}}_B}{\text{) = }}\dfrac{{number{\text{ }}of{\text{ }}moles{\text{ }}of{\text{ B}}({n_B})}}{{total\ number\ of\ moles({n_A} + {n_B})}}$
${\text{ = }}\dfrac{{5.556}}{{(0.714 + 5.556)}} = 0.886$
Step 3: To calculate the vapour pressure of pure A, we need to make use Roult’s law. According to the Roult’s law, the product of the vapour pressure of the liquid and its mole fraction equals to the partial pressure of that liquid in its solution.
Using Raoult's law, we will first find out the vapour pressure of liquid A in the solution.
As per Roult’s law, vapour pressure of pure liquid B is given by,
$partial{\text{ }}pressure{\text{ }}of{\text{ }}pure{\text{ }}liquid{\text{ }}B = vapour{\text{ }}pressure{\text{ }}of{\text{ }}liquid{\text{ }}B \times mole{\text{ }}fraction{\text{ }}of{\text{ }}liquid{\text{ }}B \\\ {p_B} = {p^0}_B \times {X_B}$
On putting the values in the above formula, we get
${p_B} = 500 \times 0.886 = 443torr$
The total vapour pressure of the solution is 475 and as per formula,
$[{P_T} = {p_A} + {p_B}]$
Substitute the respective values to find out the vapour pressure of liquid A in solution.
$\therefore {p_A} = {P_T} - {p_B} = 475 - 443 = 32torr$
The vapour pressure of liquid A in the solution is 32 torr.
Step 4: Now, to find out the vapour pressure of pure liquid A, we need to use Roult’s law.
As stated by Roult’s law, the vapour pressure of pure liquid A can be derived by,
$partial{\text{ }}pressure{\text{ }}of{\text{ }}pure{\text{ }}liquid{\text{ }}A = vapour{\text{ }}pressure{\text{ }}of{\text{ }}liquid{\text{ }}A \times mole{\text{ }}fraction{\text{ }}of{\text{ }}liquid{\text{ }}A \\\\{p_A} = {p^0}_A \times {X_A}$
Put the values in the above formula.
$32 = {p^0}_A \times 0.114$
$\therefore {p^0}_A = \dfrac{{32}}{{0.114}} = 280.7torr$

The vapour pressure of pure liquid A is 280.6torr.

Note: The vapour pressure of a liquid is its characteristic property based on strength of intermolecular bonding in that liquid. A liquid with strong intermolecular forces will have low vapour pressure and vice-versa.
The mole fraction of B is calculated using the following formula.
$[{\text{mole fraction of liquid B(}}{{\text{X}}_B}{\text{) = }}\dfrac{{number{\text{ }}of{\text{ }}moles{\text{ }}of{\text{ B}}({n_B})}}{{total{\text{ number of moles}}({n_A} + {n_B})}}]$
Alternatively, mole fraction of B can be calculated as shown below.
$[{\text{mole fraction of liquid B(}}{{\text{X}}_B}{\text{) = 1 - mole fraction of A}}]$