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What is the freezing point of a solution containing 8.1 g HBr in 100 g water assuming the acid to be 90% ionised (Kf for water=1.86 K kg mol-1) 
 
  • a)
    0.85 ºC
  • b)
    –3.53 ºC
  • c)
    0ºC
  • d)
    – 0.35ºC
Correct answer is option 'B'. Can you explain this answer?
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First, we need to calculate the molality of the solution:

Molality = moles of solute / mass of solvent in kg

Assuming 90% ionization, we can calculate the moles of HBr:

Moles of HBr = 8.1 g / 81.39 g/mol = 0.1 mol

However, since HBr is 90% ionized, the actual concentration of HBr is only 0.1 mol x 0.1 = 0.01 mol.

So, the molality is:

Molality = 0.01 mol / 0.1 kg = 0.1 mol/kg

Next, we can calculate the freezing point depression using the formula:

ΔTf = Kf x molality

where Kf is the freezing point depression constant for water (1.86 K kg mol-1).

ΔTf = 1.86 K kg mol-1 x 0.1 mol/kg = 0.186 K

Finally, we can calculate the freezing point of the solution using the formula:

Freezing point = Freezing point of pure solvent - ΔTf

Assuming the freezing point of pure water is 0°C, the freezing point of the solution is:

Freezing point = 0°C - 0.186°C = -0.186°C

Rounded to two decimal places, the freezing point of the solution is -0.19°C.
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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

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

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What is the freezing point of a solution containing 8.1 g HBr in 100 g water assuming the acid to be 90% ionised (Kffor water=1.86 K kg mol-1)a)0.85 ºCb)–3.53 ºCc)0ºCd)– 0.35ºCCorrect answer is option 'B'. Can you explain this answer?
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