Question

Two liquids A and B form an ideal solution. At $\text{ 300 K }$, the vapour pressure of a solution containing mole of A and 3 moles of B is 550 mm of Hg. At the same temperature, if one mole of B is added to this solution. The vapour pressure of the solution increases by 10 mm of Hg. Determine the vapour pressure of A and B in their pure states.

Answer 1

Hint: According to Raoult's law, the vapour pressure of a binary solution which is made of $\text{ }{{\text{n}}_{\text{A}}}\text{ }$ moles of liquid A and $\text{ }{{\text{n}}_{\text{B }}}\text{ }$moles of liquid B having partial pressure $\text{ }{{\text{p}}_{\text{A}}}\text{ }$ and $\text{ }{{\text{p}}_{\text{B }}}$ respectively is given as,
$\text{ P = }{{\text{p}}_{\text{A }}}\text{+ }{{\text{p}}_{\text{B }}}=\text{ }{{\text{X}}_{\text{A}}}\text{.p}_{\text{A}}^{0}\text{ + }{{\text{X}}_{\text{B}}}\text{.p}_{\text{B}}^{0}\text{ }$
Where, $\text{ }{{\text{X}}_{\text{A}}}$ is the mole fraction of pure A liquid and \[\text{ }{{\text{X}}_{\text{B }}}\text{ }\]is the mole fraction of pure B liquid in solution

Complete step by step answer:
We are provided with the following data:
Part a) Vapour pressure of solution is $\text{ 550 mm }$of Hg
Number of moles of A is 1 mole
Number of moles of B is 3 moles
Part b) vapour pressure of solution is increased by 10 mm of Hg
Number of moles of A is 1 mole
Number of moles of B is 3+1 i.e. 4 moles
Part a) Let the vapour pressure of pure A as the $\text{ p}_{\text{A}}^{\text{0}}\text{ }$ and let the vapour pressure of pure B as $\text{ p}_{\text{B}}^{\text{0}}\text{ }$
The total vapour pressure of the ideal solution of A and B where 1 mole of A and 3 moles of B is given as follows,
$\text{ Total vapor pressure = }{{\text{X}}_{\text{A}}}\text{ }\text{. p}_{\text{A}}^{\text{0}}\text{ }+\text{ }{{\text{X}}_{\text{B}}}\text{ }\text{. p}_{\text{B}}^{\text{0}}\text{ }$
Where, $\text{ }{{\text{X}}_{\text{A}}}$ is the mole fraction of pure A liquid and \[\text{ }{{\text{X}}_{\text{B }}}\text{ }\]is the mole fraction of pure B liquid in solution.
We know that , the vapour pressure of solution is $\text{ 550 mm }$ of mercury and mole fraction of A is ,
$\text{ }{{\text{X}}_{\text{A}}}\text{ = }\dfrac{1}{1+3}\text{ = }\dfrac{1}{4}\text{ }$
And the mole fraction of B is ,
$\text{ }{{\text{X}}_{\text{B}}}\text{ = }\dfrac{3}{1+3}\text{ = }\dfrac{3}{4}\text{ }$
The total vapour is wittren as,
\[\text{ 550 = }\dfrac{1}{4}\text{ p}_{\text{A}}^{\text{0}}\text{ }+\text{ }\dfrac{3}{4}\text{ p}_{\text{B}}^{\text{0}}\text{ }\]
Therefore, $\text{ 2200 = p}_{\text{A}}^{\text{0}}\text{ + 3 p}_{\text{B}}^{\text{0}}\text{ }$ (1)
Part b) At the same temperature, when we add one mole B to the solution ,the total number of moles of B in the solution becomes equal to 4.therefore ,the mole fraction of A becomes ,
$\text{ }{{\text{X}}_{\text{A}}}\text{ = }\dfrac{1}{1+4}\text{ = }\dfrac{1}{5}\text{ }$
And the mole fraction of B is given as,
$\text{ }{{\text{X}}_{\text{B}}}\text{ = }\dfrac{4}{1+4}\text{ = }\dfrac{4}{5}\text{ }$
The total vapour pressure when one mole of added to the B becomes equal to the vapour pressure increased by the 10 mm of Hg .thus, the final vapour pressure is given as,
$\text{ Total vapour pressure = (550 + 10) mm of Hg = 560 mm of Hg }$
The equation of vapour pressure of $\text{ p}_{\text{A}}^{\text{0}}\text{ }$and $\text{ p}_{\text{B}}^{\text{0}}\text{ }$with the vapour pressure 510 mm of Hg is given as follows,
$\text{ 560 = }\dfrac{1}{5}\text{ p}_{\text{A}}^{\text{0}}\text{ }+\text{ }\dfrac{4}{5}\text{ p}_{\text{B}}^{\text{0}}\text{ }$
Therefore, $\text{ 2800 = p}_{\text{A}}^{\text{0}}\text{ + 4 p}_{\text{B}}^{\text{0}}\text{ }$ (2)
Subtract the equation (1) from the equation (2), we get the value of vapour pressure of B as follows,
$\text{ }\begin{matrix}
\text{ 2800 = p}_{\text{A}}^{\text{0}}\text{ + 4 p}_{\text{B}}^{\text{0}}\text{ } \\
\text{ }-\text{2200 = p}_{\text{A}}^{\text{0}}\text{ + 3 p}_{\text{B}}^{\text{0}}\text{ } \\
\overline{\text{ 600 = p}_{\text{B}}^{\text{0}}\text{ }} \\
\end{matrix}\text{ }$
Thus the vapour pressure of pure B liquid is 600 mm of Hg.
Now, substitute the value of vapour pressure of B in equation (1) we have,
$\begin{align}
& \text{ 2200 = p}_{\text{A}}^{\text{0}}\text{ + 3 (600) } \\
& \text{ }\Rightarrow \text{p}_{\text{A}}^{\text{0}}\text{ = 2200 }-\text{ 1800 } \\
& \therefore \text{p}_{\text{A}}^{\text{0}}\text{ = 400 mm of Hg } \\
\end{align}$

Thus, the vapour pressure of pure A is 400 mm of Hg.

Note: Only ideal solutions obeys Raoult's law. To solve this question use the linear equation method. Add or subtract the equation in such a manner that the one of the variables gets eliminated from the equation. This will give the value of one variable in the equation. This method is known as the simultaneous equation.