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0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If Kf for water is 1.86 K kg mol– 1,the lowering in freezing point of the solution is
- a)0.56 K
- b)1.12 K [2007]
- c)– 0.56 K
- d)– 1.12 K
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If Kf f...
As ΔTf = iKfm
For
Total no. of moles = 1 – 0.20 + 0.20 + 0.20 = 1 + 0.20 = 1.2
∴ ΔTf = 1.2 × 1.86 × 0.5 = 1.1160 ≈ 1.12 K
∴ ΔTf = 1.2 × 1.86 × 0.5 = 1.1160 ≈ 1.12 K
Most Upvoted Answer
0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If Kf f...
To solve this problem, we need to first calculate the concentration of the weak acid (HX) in the solution.
A 0.5 molal solution means that there are 0.5 moles of solute (HX) dissolved in 1 kg of solvent (water). Since the solvent is water, the mass of the solution is equal to its volume (1 L = 1 kg).
Therefore, the concentration of HX in the solution is 0.5 mol/L.
Now, we need to determine the degree of ionization (α) of the weak acid in the solution. The problem states that it is 20% ionized, which means that only 20% of the weak acid molecules dissociate into ions.
So, α = 0.20.
Next, we can calculate the concentration of the ions (H+ and X-) in the solution.
Since the concentration of HX in the solution is 0.5 mol/L and only 20% of it dissociates, the concentration of H+ and X- ions is 0.5 mol/L * 0.20 = 0.10 mol/L.
Now, we can calculate the molality (m) of the solution.
Molality is defined as the number of moles of solute per kilogram of solvent.
In this case, the number of moles of solute (HX) is 0.5 moles and the mass of the solvent (water) is 1 kg.
Therefore, the molality of the solution is 0.5 mol / 1 kg = 0.5 mol/kg.
Finally, we can calculate the freezing point depression (ΔTf) using the formula:
ΔTf = Kf * m
where Kf is the freezing point depression constant for water, and m is the molality of the solution.
Given that Kf for water is 1.86 K kg mol, and the molality of the solution is 0.5 mol/kg, we can calculate:
ΔTf = 1.86 K kg mol * 0.5 mol/kg = 0.93 K
Therefore, the freezing point of the solution is depressed by 0.93 K.
A 0.5 molal solution means that there are 0.5 moles of solute (HX) dissolved in 1 kg of solvent (water). Since the solvent is water, the mass of the solution is equal to its volume (1 L = 1 kg).
Therefore, the concentration of HX in the solution is 0.5 mol/L.
Now, we need to determine the degree of ionization (α) of the weak acid in the solution. The problem states that it is 20% ionized, which means that only 20% of the weak acid molecules dissociate into ions.
So, α = 0.20.
Next, we can calculate the concentration of the ions (H+ and X-) in the solution.
Since the concentration of HX in the solution is 0.5 mol/L and only 20% of it dissociates, the concentration of H+ and X- ions is 0.5 mol/L * 0.20 = 0.10 mol/L.
Now, we can calculate the molality (m) of the solution.
Molality is defined as the number of moles of solute per kilogram of solvent.
In this case, the number of moles of solute (HX) is 0.5 moles and the mass of the solvent (water) is 1 kg.
Therefore, the molality of the solution is 0.5 mol / 1 kg = 0.5 mol/kg.
Finally, we can calculate the freezing point depression (ΔTf) using the formula:
ΔTf = Kf * m
where Kf is the freezing point depression constant for water, and m is the molality of the solution.
Given that Kf for water is 1.86 K kg mol, and the molality of the solution is 0.5 mol/kg, we can calculate:
ΔTf = 1.86 K kg mol * 0.5 mol/kg = 0.93 K
Therefore, the freezing point of the solution is depressed by 0.93 K.
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0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If Kf for water is 1.86 K kg mol– 1,the lowering in freezing point of the solution isa)0.56 Kb)1.12 K [2007]c)– 0.56 Kd)– 1.12 KCorrect answer is option 'B'. Can you explain this answer?
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0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If Kf for water is 1.86 K kg mol– 1,the lowering in freezing point of the solution isa)0.56 Kb)1.12 K [2007]c)– 0.56 Kd)– 1.12 KCorrect answer is option 'B'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about 0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If Kf for water is 1.86 K kg mol– 1,the lowering in freezing point of the solution isa)0.56 Kb)1.12 K [2007]c)– 0.56 Kd)– 1.12 KCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If Kf for water is 1.86 K kg mol– 1,the lowering in freezing point of the solution isa)0.56 Kb)1.12 K [2007]c)– 0.56 Kd)– 1.12 KCorrect answer is option 'B'. Can you explain this answer?.
0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If Kf for water is 1.86 K kg mol– 1,the lowering in freezing point of the solution isa)0.56 Kb)1.12 K [2007]c)– 0.56 Kd)– 1.12 KCorrect answer is option 'B'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about 0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If Kf for water is 1.86 K kg mol– 1,the lowering in freezing point of the solution isa)0.56 Kb)1.12 K [2007]c)– 0.56 Kd)– 1.12 KCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If Kf for water is 1.86 K kg mol– 1,the lowering in freezing point of the solution isa)0.56 Kb)1.12 K [2007]c)– 0.56 Kd)– 1.12 KCorrect answer is option 'B'. Can you explain this answer?.
Solutions for 0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If Kf for water is 1.86 K kg mol– 1,the lowering in freezing point of the solution isa)0.56 Kb)1.12 K [2007]c)– 0.56 Kd)– 1.12 KCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12.
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