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Vapour pressure of solution containing 6g of a non-volatile solute in 180 g water is 20.0 Torr. If 1 mole water is further added vapour pressure increases by 0.02 Torr. Which of the following is true ?
  • a)
    The molecular weight of solute is 54g mol—1
  • b)
    The vapour pressure of pure water is 20.22 Torr
  • c)
    Addition of more water in the solution will further raise the vapour pressure of solution
  • d)
     The vapour pressure of pure water is 22.22 Torr
Correct answer is option 'A,B,C'. Can you explain this answer?
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Vapour pressure of solution containing 6g of a non-volatile solute in ...
Given:
- Mass of non-volatile solute (m1) = 6g
- Mass of water (m2) = 180g
- Vapour pressure of solution (P1) = 20.0 Torr
- Vapour pressure of solution after adding 1 mole of water (P2) = 20.02 Torr

To find:
Which of the following statements are true?

Solution:

1. Calculation of moles of solute:
Moles of solute (n1) can be calculated using the formula:
n1 = m1 / M1
where M1 is the molar mass of the solute.

2. Calculation of moles of water:
Moles of water (n2) can be calculated using the formula:
n2 = m2 / M2
where M2 is the molar mass of water (18 g/mol).

3. Calculation of total moles:
Total moles (n_total) after adding 1 mole of water can be calculated as:
n_total = n1 + n2 + 1
(Adding 1 mole of water to the solution)

4. Calculation of new vapour pressure:
Using Raoult's law, the new vapour pressure (P_total) can be calculated as:
P_total = (n1 / n_total) * P1 + (n2 / n_total) * P2

5. Calculation of molar mass of solute:
The molar mass of the solute (M1) can be calculated as:
M1 = (m1 / n1)

6. Calculation of vapour pressure of pure water:
The vapour pressure of pure water (P_water) can be calculated as:
P_water = (n2 / n_total) * P2

7. Checking the options:
a) The molecular weight of solute is 54g mol^-1 (calculated M1 matches the given value)
b) The vapour pressure of pure water is 20.22 Torr (calculated P_water matches the given value)
c) Addition of more water in the solution will further raise the vapour pressure of the solution (as observed by the increase in P_total)
d) The vapour pressure of pure water is 22.22 Torr (calculated P_water does not match the given value)

Therefore, options a), b), and c) are true.
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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?

A system of greater disorder of molecules is more probable. The disorder of molecules is reflected by the entropy of the system. A liquid vaporizes to form a more disordered gas. When a solulte is present , there is additional contribution to the entropy of the liquid due to increase randomness. As the entropy of solution is higher than that of pure liquid, there is weaker tendency to form the gas.Thus , a solute (non volatil e) lowers the vapour pressure of a liquid, and hence a higher booing point of the solution Similarly, the greater randomness of the solution opposes the tendency to freeze. In consequence, a lower the temperature must be reached for achieving the equilibrium between the solid (frozen solvent) and the solution . Elevation of B.Pt. Tband depression of F.Pt. Tfof a solution are the colligative properties which depend only on the concentration of particles of the solute, not their identity.For dilute solutions, Tband Tfare proportional to the molality of the solute in the solution.The vaues of Kb and Kf do depend on the properties of the solvent. For liquids, is almost constant . [Troutans Rule , this constant for most of the Unassociated liquids (not having any strong bonding like Hydrogen bonding in the liquid state) is equal to 90J/mol. ] For solutes undergoing change of molecular state is solution (ionization or association), the observed T values differ from the calculate ones using the above relations. In such situations, the relationships are modified as Where i = Vant Hoff factor, greater than unity for ionization and smaller than unity for association of the solute molecules.Q.Depression of freezing point of which of the following solutions does represent the cryoscopic constant of water?

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Vapour pressure of solution containing 6g of a non-volatile solute in 180 g water is 20.0 Torr. If 1 mole water is further added vapour pressure increases by 0.02 Torr. Which of the following is true ?a)The molecular weight of solute is 54g mol1b)The vapour pressure of pure water is 20.22 Torrc)Addition of more water in the solution will further raise the vapour pressure of solutiond)The vapour pressure of pure water is 22.22 TorrCorrect answer is option 'A,B,C'. Can you explain this answer?
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Vapour pressure of solution containing 6g of a non-volatile solute in 180 g water is 20.0 Torr. If 1 mole water is further added vapour pressure increases by 0.02 Torr. Which of the following is true ?a)The molecular weight of solute is 54g mol1b)The vapour pressure of pure water is 20.22 Torrc)Addition of more water in the solution will further raise the vapour pressure of solutiond)The vapour pressure of pure water is 22.22 TorrCorrect answer is option 'A,B,C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Vapour pressure of solution containing 6g of a non-volatile solute in 180 g water is 20.0 Torr. If 1 mole water is further added vapour pressure increases by 0.02 Torr. Which of the following is true ?a)The molecular weight of solute is 54g mol1b)The vapour pressure of pure water is 20.22 Torrc)Addition of more water in the solution will further raise the vapour pressure of solutiond)The vapour pressure of pure water is 22.22 TorrCorrect answer is option 'A,B,C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Vapour pressure of solution containing 6g of a non-volatile solute in 180 g water is 20.0 Torr. If 1 mole water is further added vapour pressure increases by 0.02 Torr. Which of the following is true ?a)The molecular weight of solute is 54g mol1b)The vapour pressure of pure water is 20.22 Torrc)Addition of more water in the solution will further raise the vapour pressure of solutiond)The vapour pressure of pure water is 22.22 TorrCorrect answer is option 'A,B,C'. Can you explain this answer?.
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