
Two liquids A and B form an ideal solution. At 300K the vapour pressure of a solution of 1 mole of A and x moles of B is 550mm. If the vapour pressures of pure A and B are 400mm and 600mm respectively, then x is:
A) 1
B) 2
C) 3
D) 4
Answer
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Hint: The answer to this question is based on the Raoult’s law which measures the vapour pressure of the solution and the calculation based on its formula for two ideal solution which is given by,$P={{\chi }_{A}}{{P}_{A}}^{0}+{{\chi }_{B}}{{P}_{B}}^{0}$
Complete answer:
We have come across the concepts of the physical chemistry part that deals with the colligative properties of the solutions.
Now let us see what Raoult's law states and how vapour pressure can be calculated.
- Raoult’s law suggests that the difference between the vapour pressure of the pure solvent and that of solution increases as the mole fraction of the solvent decreases.
- Mole fraction plays a role in this where it is defined as the number of moles of that component in that component in that particular phase.
- Now, according to the data given vapour pressure of liquid A is 400mm and vapour pressure of liquid B is 600 mm.
That is${{P}_{A}}^{0}$= 400 mm and ${{P}_{B}}^{0}$= 600 mm
Thus, total vapour pressure of the ideal solution according to Raoult’s law is as follows,
$P={{\chi }_{A}}{{P}_{A}}^{0}+{{\chi }_{B}}{{P}_{B}}^{0}$ ……..(1)
where, ${{\chi }_{A}}$and ${{\chi }_{B}}$ are mole fraction of solution A and B respectively.
Now, since there is 1 mole of A and let x be the number of moles B then total moles will be 1 + x
Now substituting all these values in equation (1), we get
$550=\dfrac{1}{1+x}\left( 400 \right)+\dfrac{x}{1+x}\left( 600 \right)$
Now rearranging this above equation we get,
\[550(1+x)=400+600x\]
\[600x-550x=550-400\]
\[\Rightarrow 50x=150\]
\[\Rightarrow x=3\]
Thus, the correct answer is the value of x = 3.
Note:
Note that the Raoult’s law is most widely applicable for the non – aqueous mixtures and is used for the estimation of contribution of individual components of a liquid or solid mixture to the total pressure exerted by the system.
Complete answer:
We have come across the concepts of the physical chemistry part that deals with the colligative properties of the solutions.
Now let us see what Raoult's law states and how vapour pressure can be calculated.
- Raoult’s law suggests that the difference between the vapour pressure of the pure solvent and that of solution increases as the mole fraction of the solvent decreases.
- Mole fraction plays a role in this where it is defined as the number of moles of that component in that component in that particular phase.
- Now, according to the data given vapour pressure of liquid A is 400mm and vapour pressure of liquid B is 600 mm.
That is${{P}_{A}}^{0}$= 400 mm and ${{P}_{B}}^{0}$= 600 mm
Thus, total vapour pressure of the ideal solution according to Raoult’s law is as follows,
$P={{\chi }_{A}}{{P}_{A}}^{0}+{{\chi }_{B}}{{P}_{B}}^{0}$ ……..(1)
where, ${{\chi }_{A}}$and ${{\chi }_{B}}$ are mole fraction of solution A and B respectively.
Now, since there is 1 mole of A and let x be the number of moles B then total moles will be 1 + x
Now substituting all these values in equation (1), we get
$550=\dfrac{1}{1+x}\left( 400 \right)+\dfrac{x}{1+x}\left( 600 \right)$
Now rearranging this above equation we get,
\[550(1+x)=400+600x\]
\[600x-550x=550-400\]
\[\Rightarrow 50x=150\]
\[\Rightarrow x=3\]
Thus, the correct answer is the value of x = 3.
Note:
Note that the Raoult’s law is most widely applicable for the non – aqueous mixtures and is used for the estimation of contribution of individual components of a liquid or solid mixture to the total pressure exerted by the system.
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