Which of the following solutions can form a minimum boiling azeotrope?
Show Hint
- \( \text{C}_2\text{H}_5\text{OH} + \text{H}_2\text{O} \)
- \( \text{n-heptane} + \text{n-hexane} \)
- \( \text{CH}_3\text{COOH} + \text{C}_5\text{H}_5\text{N} \)
- \( \text{C}_2\text{H}_5\text{Br} + \text{C}_2\text{H}_5\text{I} \)
The Correct Option is A
Solution and Explanation
A minimum boiling azeotrope is formed when the mixture exhibits a boiling point lower than that of either pure component. Examine the components of the given mixtures:
Ethanol (C\(_2\)H\(_5\)OH) and Water (H\(_2\)O) form a minimum boiling azeotrope at about 78.2°C.
This azeotrope cannot be separated by distillation, making it the correct option. The other mixtures like n-heptane and n-hexane form ideal solutions and do not form minimum boiling azeotropes. Thus, the correct answer is (1).
Top Questions on Solutions
- An aqueous solution of 0.1 M HA shows depression in freezing point of 0.2°C. If $ K_f $ (H$_2$O) = 1.86 K kg mol$^{-1}$ and assuming molarity = molality, find the dissociation constant of HA.
- Concentrated nitric acid is labelled as 75%by mass. The volume in mL of the solution which contains 30 g of nitric acid is: Given: Density of nitric acid solution is 1.25 g/mL.
- Arrange the following in increasing order of solubility product: \[ {Ca(OH)}_2, {AgBr}, {PbS}, {HgS} \]
- Density of 3 M NaCl solution is 1.25 g/mL. The molality of the solution is:
- 20 mL of 2 M NaOH solution is added to 400 mL of 0.5 M NaOH solution. The final concentration of the solution is \(\_\_\_\_\_\) x \(10^{-2}\)M. (Nearest integer)
Questions Asked in JEE Main exam
- Number of solutions in the interval \( [-2\pi, 2\pi] \) for the equation: \[ 2\sqrt{2} \cos^2 \theta + (2 - \sqrt{6}) \cos \theta - \sqrt{3} = 0 \]
- JEE Main - 2025
- Trigonometry
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R
Assertion A: Work done in moving a test charge between two points inside a uniformly charged spherical shell is zero, no matter which path is chosen.
Reason R: Electrostatic potential inside a uniformly charged spherical shell is constant and is same as that on the surface of the shell.
In the light of the above statements, choose the correct answer from the options given below- JEE Main - 2025
- Electrostatics
- Water falls from a height of 200 m into a pool. Calculate the rise in temperature of the water assuming no heat dissipation from the water in the pool. (Take $ g = 10 \, \text{m/s}^2 $, specific heat of water = 4200 J/(kg K))
- JEE Main - 2025
- Thermodynamics
Figure shows a current carrying square loop ABCD of edge length is $ a $ lying in a plane. If the resistance of the ABC part is $ r $ and that of the ADC part is $ 2r $, then the magnitude of the resultant magnetic field at the center of the square loop is:
- JEE Main - 2025
- Electromagnetism
The output voltage in the following circuit is (Consider ideal diode case):
- JEE Main - 2025
- Zener Diodes