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The freezing point depression of a 1.00x 10-3 molal aqueous solution of Kx[Fe(CN)6] is 7.1O x 10-3 K. If Kf (H2O)= 1.86 K kg mol-1, what is the value of x?
Correct answer is '3'. Can you explain this answer?
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The freezing point depression of a 1.00x 10-3molal aqueous solution of...
∴ i = van't Hoff factor = 1 + ( y - 1 )α
where, α = degree of ionisation = 1 = 1 + (x + 1 — 1) 1 = (x + 1)
ΔTf = molality x Kf x i
7 x 10-3 = 1 x 10-3 x 1.86 x (x + 1)
x + 1 = 3.76 = 4
∴ x = 3
Most Upvoted Answer
The freezing point depression of a 1.00x 10-3molal aqueous solution of...
Given:
- Molality (m) of the solution = 1.00×10-3 mol/kg
- Freezing point depression (∆Tf) = 7.1×10-3 K
- Freezing point depression constant (Kf) of water = 1.86 K kg/mol
To find:
- The value of x in the compound Kx[Fe(CN)6]
Solution:
Step 1: Calculate the moles of solute (Kx[Fe(CN)6]):
- We are given the molality of the solution, which is defined as the number of moles of solute per kilogram of solvent.
- Molality (m) = moles of solute (Kx[Fe(CN)6]) / mass of solvent (water in this case)
- We assume 1 kg of water, so the moles of solute can be calculated as:
- Moles of solute (Kx[Fe(CN)6]) = molality (m) × mass of solvent (water)
- Moles of solute (Kx[Fe(CN)6]) = 1.00×10-3 mol/kg × 1 kg
- Moles of solute (Kx[Fe(CN)6]) = 1.00×10-3 mol
Step 2: Calculate the molality of the solute (Kx[Fe(CN)6]):
- Molality (m) = moles of solute (Kx[Fe(CN)6]) / mass of solvent (water)
- We can rearrange this equation to find the mass of solvent:
- Mass of solvent (water) = moles of solute (Kx[Fe(CN)6]) / molality (m)
- Mass of solvent (water) = 1.00×10-3 mol / 1.00×10-3 mol/kg
- Mass of solvent (water) = 1 kg
Step 3: Calculate the molal freezing point depression constant:
- The molal freezing point depression constant (Kf) is defined as the change in freezing point per mole of solute per kilogram of solvent.
- ∆Tf = Kf × m
- We can rearrange this equation to find the molal freezing point depression constant:
- Kf = ∆Tf / m
- Kf = 7.1×10-3 K / 1.00×10-3 mol/kg
- Kf = 7.1 K
Step 4: Calculate the value of x:
- The freezing point depression is caused by the dissociation of Kx[Fe(CN)6] into K+ and [Fe(CN)6]- ions.
- Each mole of Kx[Fe(CN)6] dissociates into x moles of K+ and 1 mole of [Fe(CN)6]-
- Therefore, the change in freezing point (∆Tf) is proportional to the number of moles of particles formed.
- Since the freezing point depression is 7.1 K and the molal freezing point depression constant (Kf) is 7
- Molality (m) of the solution = 1.00×10-3 mol/kg
- Freezing point depression (∆Tf) = 7.1×10-3 K
- Freezing point depression constant (Kf) of water = 1.86 K kg/mol
To find:
- The value of x in the compound Kx[Fe(CN)6]
Solution:
Step 1: Calculate the moles of solute (Kx[Fe(CN)6]):
- We are given the molality of the solution, which is defined as the number of moles of solute per kilogram of solvent.
- Molality (m) = moles of solute (Kx[Fe(CN)6]) / mass of solvent (water in this case)
- We assume 1 kg of water, so the moles of solute can be calculated as:
- Moles of solute (Kx[Fe(CN)6]) = molality (m) × mass of solvent (water)
- Moles of solute (Kx[Fe(CN)6]) = 1.00×10-3 mol/kg × 1 kg
- Moles of solute (Kx[Fe(CN)6]) = 1.00×10-3 mol
Step 2: Calculate the molality of the solute (Kx[Fe(CN)6]):
- Molality (m) = moles of solute (Kx[Fe(CN)6]) / mass of solvent (water)
- We can rearrange this equation to find the mass of solvent:
- Mass of solvent (water) = moles of solute (Kx[Fe(CN)6]) / molality (m)
- Mass of solvent (water) = 1.00×10-3 mol / 1.00×10-3 mol/kg
- Mass of solvent (water) = 1 kg
Step 3: Calculate the molal freezing point depression constant:
- The molal freezing point depression constant (Kf) is defined as the change in freezing point per mole of solute per kilogram of solvent.
- ∆Tf = Kf × m
- We can rearrange this equation to find the molal freezing point depression constant:
- Kf = ∆Tf / m
- Kf = 7.1×10-3 K / 1.00×10-3 mol/kg
- Kf = 7.1 K
Step 4: Calculate the value of x:
- The freezing point depression is caused by the dissociation of Kx[Fe(CN)6] into K+ and [Fe(CN)6]- ions.
- Each mole of Kx[Fe(CN)6] dissociates into x moles of K+ and 1 mole of [Fe(CN)6]-
- Therefore, the change in freezing point (∆Tf) is proportional to the number of moles of particles formed.
- Since the freezing point depression is 7.1 K and the molal freezing point depression constant (Kf) is 7
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Community Answer
The freezing point depression of a 1.00x 10-3molal aqueous solution of...
∴ i = van't Hoff factor = 1 + ( y - 1 )α
where, α = degree of ionisation = 1 = 1 + (x + 1 — 1) 1 = (x + 1)
ΔTf = molality x Kf x i
7 x 10-3 = 1 x 10-3 x 1.86 x (x + 1)
x + 1 = 3.76 = 4
∴ x = 3
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The freezing point depression of a 1.00x 10-3molal aqueous solution of Kx[Fe(CN)6] is 7.1O x 10-3 K. If Kf (H2O)= 1.86 K kg mol-1, what is the value of x?Correct answer is '3'. Can you explain this answer?
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The freezing point depression of a 1.00x 10-3molal aqueous solution of Kx[Fe(CN)6] is 7.1O x 10-3 K. If Kf (H2O)= 1.86 K kg mol-1, what is the value of x?Correct answer is '3'. 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 The freezing point depression of a 1.00x 10-3molal aqueous solution of Kx[Fe(CN)6] is 7.1O x 10-3 K. If Kf (H2O)= 1.86 K kg mol-1, what is the value of x?Correct answer is '3'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The freezing point depression of a 1.00x 10-3molal aqueous solution of Kx[Fe(CN)6] is 7.1O x 10-3 K. If Kf (H2O)= 1.86 K kg mol-1, what is the value of x?Correct answer is '3'. Can you explain this answer?.
The freezing point depression of a 1.00x 10-3molal aqueous solution of Kx[Fe(CN)6] is 7.1O x 10-3 K. If Kf (H2O)= 1.86 K kg mol-1, what is the value of x?Correct answer is '3'. 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 The freezing point depression of a 1.00x 10-3molal aqueous solution of Kx[Fe(CN)6] is 7.1O x 10-3 K. If Kf (H2O)= 1.86 K kg mol-1, what is the value of x?Correct answer is '3'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The freezing point depression of a 1.00x 10-3molal aqueous solution of Kx[Fe(CN)6] is 7.1O x 10-3 K. If Kf (H2O)= 1.86 K kg mol-1, what is the value of x?Correct answer is '3'. Can you explain this answer?.
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