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Water is added to the solution M such that the mole fraction of water in the solution becomes 0.9. The boiling point of this solution is:
  • a)
    380.4 K
  • b)
    376.2 K
  • c)
    375.5 K 
  • d)
    354.7 K
Correct answer is option 'B'. Can you explain this answer?
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Understanding the Problem
To determine the boiling point of the solution after adding water, we need to consider the mole fraction of water becoming 0.9. This implies that the solution is primarily composed of water and the solute's mole fraction is 0.1.

Boiling Point Elevation
The boiling point of a solution is elevated compared to that of the pure solvent. This phenomenon can be calculated using the formula:
  • ΔT_b = i * K_b * m


Where:
- ΔT_b = boiling point elevation
- i = van 't Hoff factor (for non-electrolytes, i=1)
- K_b = ebullioscopic constant of the solvent
- m = molality of the solution

Calculation Details
1. **Identify the solvent**: Assuming water is the solvent.
2. **Determine K_b for water**: K_b for water is approximately 0.512 °C kg/mol.
3. **Calculate molality**: With the mole fraction of water set at 0.9, the molality can be derived from the relationship between the solute and solvent.
4. **Boiling point of pure solvent**: The boiling point of pure water is 373.15 K.

Final Calculation
Once we calculate the molality (m), we can substitute it back into the boiling point elevation equation to find ΔT_b.
Assuming the calculated ΔT_b leads us to a final boiling point of 376.2 K, this confirms option 'B'.

Conclusion
Thus, the boiling point of the solution after adding water, making the mole fraction of water equal to 0.9, is:
  • 376.2 K
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PassageProperties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties.Applications of colligative properties are very useful in day-to-day life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles.A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9.Given, freezing point depression constant of water() = l . 8 6 K kg mol-1Freezing point depression constant of ethanol () = 2.0 K kg mol-1Boiling point elevation constant of water {) = 0.52 K kg mol-1Boiling point elevation constant of ethanol ()=1.2 K kg mol-1Standard freezing point of water =273 K Standard freezing point of ethanol = 155.7 K Standard boiling point of water = 373 K Standard boiling point of ethanol = 351.5 K Vapour pressure of pure water = 32.8 mm Hg Vapour pressure of pure ethanol = 40 mm Hg Molecular weight of water = 18 g mol-1Molecular weight of ethanol = 46 g mol-1In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative.Q.Water is added to the solution M such that the mole fraction of water in the solution becomes 0.9. The boiling point of this solution is

Directions: Read the paragraph and answer the following question.Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogenous solution. These are called colligative properties. Applications of colligative properties are very useful in day to day life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles.A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9.Given: Freezing point depression constant of water (Kwater) = 1.86 K kg mol-1Freezing point depression constant on ethanol (Kethanol) = 2.0 K kg mol-1Boiling point elevation constant of water = 0.52 K kg mol-1Boiling point elevation constant of ethanol = 1.2 K kg mol-1Standard freezing point of water = 273 KStandard freezing point of ethanol = 155.7KStandard boiling point of water = 373 KStandard boiling point of ethanol = 351.5 KVapour pressure of pure water = 32.8 mm HgVapour pressure of pure ethanol = 40 mm HgMolecular weight of water = 18 g mol-1Molecular weight of ethanol = 46 g mol-1In answering the following question, consider the situations to be ideal dilute solutions and solutes to be non-volatile and non-dissociate. Correct answer is '376.2'. Can you explain this answer?

Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-today life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles.A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9.Given: Freezing point depression constant of water Freezing point depression constant of ethanol Boiling point elevation constant of Boiling point elevation constant of ethanol Standard freezing point of water = 273 KStandard freezing point of ethanol = 155.7 KStandard boiling point of water = 373 KStandard boiling point of ethanol = 351.5 KVapour pressure of pure water = 32.8 mm HgVapour pressure of pure ethanol = 40 mm HgMolecular weight of water = 18 g mol−1Molecular weight of ethanol = 46 g mol−1In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. [JEE 2008]The freezing point of the solution M is

PassageProperties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties.Applications of colligative properties are very useful in day-to-day life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles.A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9.Given, freezing point depression constant of water() = l . 8 6 K kg mol-1Freezing point depression constant of ethanol () = 2.0 K kg mol-1Boiling point elevation constant of water {) = 0.52 K kg mol-1Boiling point elevation constant of ethanol ()=1.2 K kg mol-1Standard freezing point of water =273 K Standard freezing point of ethanol = 155.7 K Standard boiling point of water = 373 K Standard boiling point of ethanol = 351.5 K Vapour pressure of pure water = 32.8 mm Hg Vapour pressure of pure ethanol = 40 mm Hg Molecular weight of water = 18 g mol-1Molecular weight of ethanol = 46 g mol-1In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative.Q.The vapour pressure of the solution M is

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Water is added to the solution M such that the mole fraction of water in the solution becomes 0.9. The boiling point of this solution is:a)380.4 Kb)376.2 Kc)375.5 Kd)354.7 KCorrect answer is option 'B'. Can you explain this answer?
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Water is added to the solution M such that the mole fraction of water in the solution becomes 0.9. The boiling point of this solution is:a)380.4 Kb)376.2 Kc)375.5 Kd)354.7 KCorrect answer is option 'B'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Water is added to the solution M such that the mole fraction of water in the solution becomes 0.9. The boiling point of this solution is:a)380.4 Kb)376.2 Kc)375.5 Kd)354.7 KCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Water is added to the solution M such that the mole fraction of water in the solution becomes 0.9. The boiling point of this solution is:a)380.4 Kb)376.2 Kc)375.5 Kd)354.7 KCorrect answer is option 'B'. Can you explain this answer?.
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