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0.2 molal acid HX is 20% ionised in solution. Kf = 1.86 Kmolality–1. The freezing point of the solution is : (1995S)
- a)– 0.45
- b)– 0.90
- c)– 0.31
- d)– 0.53
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
0.2 molal acid HX is 20% ionised in solution. Kf = 1.86 Kmolality&ndas...
Depression in freezing point, 
Van’t Hoff factor, 
where n = no. of ions
where n = no. of ions
[ For 20 % ionisation, α = 20/100= 0.2]
∴ΔTf = 1.2 × 1.86 × 0.2 = 0.45 [ ∵ m = 0.2]
Hence freezing point of solution is 0 – 0.45 = –0.45
[∵ F.P of water = 0.0 C]
∴ΔTf = 1.2 × 1.86 × 0.2 = 0.45 [ ∵ m = 0.2]
Hence freezing point of solution is 0 – 0.45 = –0.45
[∵ F.P of water = 0.0 C]
Most Upvoted Answer
0.2 molal acid HX is 20% ionised in solution. Kf = 1.86 Kmolality&ndas...
We can use the formula:
ΔTf = Kf * molality * i
where ΔTf is the freezing point depression, Kf is the freezing point depression constant (given as 1.86 Kmolality), molality is the concentration of the acid in mol/kg, and i is the van't Hoff factor, which represents the number of particles formed in solution per molecule of acid.
First, we need to calculate the van't Hoff factor for the acid. Since it is 20% ionized, we can assume that for every molecule of HX that dissolves, 0.2 * 0.2 = 0.04 moles of H+ and X- ions are formed. Thus, the van't Hoff factor i is:
i = (0.2 + 0.04 + 0.04) / 0.2 = 1.4
Next, we can substitute the values into the formula:
ΔTf = 1.86 * 0.2 * 1.4 = 0.5232 K
Therefore, the freezing point of the solution is lowered by 0.5232 K due to the presence of the acid.
ΔTf = Kf * molality * i
where ΔTf is the freezing point depression, Kf is the freezing point depression constant (given as 1.86 Kmolality), molality is the concentration of the acid in mol/kg, and i is the van't Hoff factor, which represents the number of particles formed in solution per molecule of acid.
First, we need to calculate the van't Hoff factor for the acid. Since it is 20% ionized, we can assume that for every molecule of HX that dissolves, 0.2 * 0.2 = 0.04 moles of H+ and X- ions are formed. Thus, the van't Hoff factor i is:
i = (0.2 + 0.04 + 0.04) / 0.2 = 1.4
Next, we can substitute the values into the formula:
ΔTf = 1.86 * 0.2 * 1.4 = 0.5232 K
Therefore, the freezing point of the solution is lowered by 0.5232 K due to the presence of the acid.
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0.2 molal acid HX is 20% ionised in solution. Kf = 1.86 Kmolality–1. The freezing point of the solution is : (1995S)a)– 0.45b)– 0.90c)– 0.31d)– 0.53Correct answer is option 'A'. Can you explain this answer?
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0.2 molal acid HX is 20% ionised in solution. Kf = 1.86 Kmolality–1. The freezing point of the solution is : (1995S)a)– 0.45b)– 0.90c)– 0.31d)– 0.53Correct answer is option 'A'. 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 0.2 molal acid HX is 20% ionised in solution. Kf = 1.86 Kmolality–1. The freezing point of the solution is : (1995S)a)– 0.45b)– 0.90c)– 0.31d)– 0.53Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 0.2 molal acid HX is 20% ionised in solution. Kf = 1.86 Kmolality–1. The freezing point of the solution is : (1995S)a)– 0.45b)– 0.90c)– 0.31d)– 0.53Correct answer is option 'A'. Can you explain this answer?.
0.2 molal acid HX is 20% ionised in solution. Kf = 1.86 Kmolality–1. The freezing point of the solution is : (1995S)a)– 0.45b)– 0.90c)– 0.31d)– 0.53Correct answer is option 'A'. 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 0.2 molal acid HX is 20% ionised in solution. Kf = 1.86 Kmolality–1. The freezing point of the solution is : (1995S)a)– 0.45b)– 0.90c)– 0.31d)– 0.53Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 0.2 molal acid HX is 20% ionised in solution. Kf = 1.86 Kmolality–1. The freezing point of the solution is : (1995S)a)– 0.45b)– 0.90c)– 0.31d)– 0.53Correct answer is option 'A'. Can you explain this answer?.
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