
If 100mL liquid A and 50mL liquid B are mixed to form $138\,mL$ solution.
A. It is an ideal solution
B. has high boiling azeotrope
C. has low boiling azeotrope
D. None of the above
Answer
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Hint: Azeotropes is defined as a mixture having two or more liquids in which proportions cannot be changed by a simple distillation process. When an azeotrope is boiled, the vapour shows the same proportion of constituent as unboiled mixture as there is no change by distillation.
Complete step by step answer:
Here, it is given that the volume of liquid A is \[100\,mL\] and volume of liquid B is $50\,mL$
On mixing liquid, A and liquid B, it is given that the volume of the solution (non – ideal) is $138\,mL$
Now, let us calculate the ideal solution
${{V}_{I.S}}={{V}_{a}}+{{V}_{b}}$
where, ${{V}_{I.S}}$ is the ideal solution, \[{{V}_{a}}\] is the volume of liquid A and ${{V}_{b}}$ is the volume of liquid B.
now, substituting the values in the above formula, we get,
${{V}_{I.S}}=100+50$
$\Rightarrow {{V}_{I.S}}=150\,mL$
Now, we will calculate the change in the volume of the solution
$\Delta {{V}_{mix}}={{V}_{N.I}}-{{V}_{I.S}}$
where, $\Delta {{V}_{mix}}$ is the change in the volume of mixture, ${{V}_{N.I}}$ is the volume of non – ideal solution and ${{V}_{I.S}}$ is the volume of ideal solution.
On, substituting the values in the above formula, we get,
$\Delta {{V}_{mix}}=138-150$
$\Rightarrow \Delta {{V}_{mix}}=-12$
As we can see that the change in the volume of the mixture is negative. This condition occurs when the vapor pressure of the solution is very less due to the strong intermolecular forces of the particles. That means, it requires more heat to weaken the force of attraction between the particles. It requires high boiling point. Hence has a high boiling point azeotrope.
So, the correct answer is Option B.
Note: In the conclusion, we can see that change in the volume is negative. The vapour pressure of the solution is less due to strong forces that means it requires more heat to weaken the force of attraction and requires high boiling point.
Boiling point is defined as the temperature at which vapor pressure of a liquid is equal to the vapor pressure of the surrounding liquid. At this temperature, liquid is converted into vapor.
Complete step by step answer:
Here, it is given that the volume of liquid A is \[100\,mL\] and volume of liquid B is $50\,mL$
On mixing liquid, A and liquid B, it is given that the volume of the solution (non – ideal) is $138\,mL$
Now, let us calculate the ideal solution
${{V}_{I.S}}={{V}_{a}}+{{V}_{b}}$
where, ${{V}_{I.S}}$ is the ideal solution, \[{{V}_{a}}\] is the volume of liquid A and ${{V}_{b}}$ is the volume of liquid B.
now, substituting the values in the above formula, we get,
${{V}_{I.S}}=100+50$
$\Rightarrow {{V}_{I.S}}=150\,mL$
Now, we will calculate the change in the volume of the solution
$\Delta {{V}_{mix}}={{V}_{N.I}}-{{V}_{I.S}}$
where, $\Delta {{V}_{mix}}$ is the change in the volume of mixture, ${{V}_{N.I}}$ is the volume of non – ideal solution and ${{V}_{I.S}}$ is the volume of ideal solution.
On, substituting the values in the above formula, we get,
$\Delta {{V}_{mix}}=138-150$
$\Rightarrow \Delta {{V}_{mix}}=-12$
As we can see that the change in the volume of the mixture is negative. This condition occurs when the vapor pressure of the solution is very less due to the strong intermolecular forces of the particles. That means, it requires more heat to weaken the force of attraction between the particles. It requires high boiling point. Hence has a high boiling point azeotrope.
So, the correct answer is Option B.
Note: In the conclusion, we can see that change in the volume is negative. The vapour pressure of the solution is less due to strong forces that means it requires more heat to weaken the force of attraction and requires high boiling point.
Boiling point is defined as the temperature at which vapor pressure of a liquid is equal to the vapor pressure of the surrounding liquid. At this temperature, liquid is converted into vapor.
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