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Freezing point of a 4% aqueous solution of X is equal to freezing point of 12% aqueous solution of Y.
If molecular weight of X is A, then molecular weight of Y is
  • a)
    2A
  • b)
    3A
  • c)
    A
  • d)
    4A
Correct answer is option 'B'. Can you explain this answer?
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Freezing point of a 4% aqueous solution of X is equal to freezing poin...
For same freezing point, molality of both solution should be same.

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Freezing point of a 4% aqueous solution of X is equal to freezing poin...
Given: Freezing point of a 4% aqueous solution of X is equal to freezing point of 12% aqueous solution of Y. Molecular weight of X is A.

To find: Molecular weight of Y

Solution:
1. Use the formula for freezing point depression:

ΔTf = Kf*m

where ΔTf is the freezing point depression, Kf is the freezing point depression constant for the solvent (water), and m is the molality of the solution (moles of solute per kg of solvent).

2. Since the freezing point of the two solutions is the same:

ΔTf(X) = ΔTf(Y)
Kf*m(X) = Kf*m(Y)

where m(X) and m(Y) are the molalities of X and Y solutions, respectively.

3. Rewrite the molality in terms of the percent concentration:

m(X) = (4 g X / 100 g solvent) / (A g/mol X) = 0.04 / A mol/kg solvent
m(Y) = (12 g Y / 100 g solvent) / (Mol. Wt. of Y g/mol Y) = 0.12 / (Mol. Wt. of Y) mol/kg solvent

4. Substitute into the equation from step 2:

Kf*0.04/A = Kf*0.12/(Mol. Wt. of Y)
Mol. Wt. of Y/A = 3

5. Therefore, the molecular weight of Y is 3 times the molecular weight of X, which is option B.
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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

Chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules (approximately 6.023 × 1023) are present in a few grams of any chemical compound varying with their atomic/molecular masses. To handle such large numbers conveniently, the mole concept was introduced.This concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry and radiochemistry. The following example illustrates a typical case, involving chemical/electrochemical reaction, which requires a clear understanding of the mole concept. A 4.0 molar aqueous solution NaCl is prepared and 500 mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrodes (atomic mass: Na = 23, Hg = 200; 1 Faraday = 96,500 coulombs)If the cathode is a Hg electrode, the maximum weight (g) of amalgam formed from this solution is Correct answer is '446'. Can you explain this answer?

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Freezing point of a 4% aqueous solution of X is equal to freezing point of 12% aqueous solution of Y.If molecular weight of X is A, then molecular weight of Y isa)2Ab)3Ac)Ad)4ACorrect answer is option 'B'. Can you explain this answer?
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