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Passage II
A solution containing 10 g of a dibasic acid in 1000 g of water freezes at -0.15° C. 10 mL of the acid is neutralised by 12 mL of 0.1 N NaOH solution. [Kf (H20 ) = 1.86° mol-1 kg]
Q.
van’t Hoff factor (/') of the dibasic acid is
- a)1.34
- b)2.00
- c)2.72
- d)3.00
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Passage IIA solution containing 10 g of a dibasic acid in 1000 g of wa...
y = 3 = number of ions from one unit
i = van't Hoff factor = 1 + (y - 1)x = (1 + 2x)
Also, 10 mL of dibasic acid = 12mL of 0.1 N NaOH
N1 (dibasic acid) = 0.12 g equivL-1
0.12 x equivalent weight = concentration = 10 g L-1'
Most Upvoted Answer
Passage IIA solution containing 10 g of a dibasic acid in 1000 g of wa...
Given:
- Mass of dibasic acid = 10 g
- Mass of water = 1000 g
- Freezing point depression = -0.15 °C
- Volume of acid neutralized = 10 mL
- Volume of NaOH solution = 12 mL
- Concentration of NaOH solution = 0.1 N
- Van 't Hoff factor (i) of the dibasic acid is to be determined.
Solution:
1. Calculate the molality of the solution using the freezing point depression formula:
ΔTf = Kf * m
Where:
ΔTf = Freezing point depression
Kf = Cryoscopic constant for water (1.86 mol-1kg)
m = Molality of the solution
Substituting the given values:
-0.15 = 1.86 * m
Solving for m:
m = -0.15 / 1.86
m ≈ -0.0806 mol/kg
2. Calculate the moles of the dibasic acid using the molality and mass of water:
Moles of solute = molality * mass of solvent (in kg)
Substituting the given values:
Moles of solute = -0.0806 * 1
Moles of solute ≈ -0.0806 mol
3. Determine the moles of the dibasic acid that reacted with NaOH:
Moles of solute reacted with NaOH = (Volume of acid neutralized / 1000) * Concentration of NaOH
Substituting the given values:
Moles of solute reacted with NaOH = (10 / 1000) * 0.1
Moles of solute reacted with NaOH = 0.001 mol
4. Calculate the Van 't Hoff factor (i) using the ratio of moles reacted with NaOH to the moles of solute:
i = Moles of solute reacted with NaOH / Moles of solute
Substituting the calculated values:
i = 0.001 / -0.0806
i ≈ -1.24
Since the Van 't Hoff factor cannot be negative, we take the absolute value:
i ≈ 1.24
Therefore, the Van 't Hoff factor (i) of the dibasic acid is approximately 1.24.
- Mass of dibasic acid = 10 g
- Mass of water = 1000 g
- Freezing point depression = -0.15 °C
- Volume of acid neutralized = 10 mL
- Volume of NaOH solution = 12 mL
- Concentration of NaOH solution = 0.1 N
- Van 't Hoff factor (i) of the dibasic acid is to be determined.
Solution:
1. Calculate the molality of the solution using the freezing point depression formula:
ΔTf = Kf * m
Where:
ΔTf = Freezing point depression
Kf = Cryoscopic constant for water (1.86 mol-1kg)
m = Molality of the solution
Substituting the given values:
-0.15 = 1.86 * m
Solving for m:
m = -0.15 / 1.86
m ≈ -0.0806 mol/kg
2. Calculate the moles of the dibasic acid using the molality and mass of water:
Moles of solute = molality * mass of solvent (in kg)
Substituting the given values:
Moles of solute = -0.0806 * 1
Moles of solute ≈ -0.0806 mol
3. Determine the moles of the dibasic acid that reacted with NaOH:
Moles of solute reacted with NaOH = (Volume of acid neutralized / 1000) * Concentration of NaOH
Substituting the given values:
Moles of solute reacted with NaOH = (10 / 1000) * 0.1
Moles of solute reacted with NaOH = 0.001 mol
4. Calculate the Van 't Hoff factor (i) using the ratio of moles reacted with NaOH to the moles of solute:
i = Moles of solute reacted with NaOH / Moles of solute
Substituting the calculated values:
i = 0.001 / -0.0806
i ≈ -1.24
Since the Van 't Hoff factor cannot be negative, we take the absolute value:
i ≈ 1.24
Therefore, the Van 't Hoff factor (i) of the dibasic acid is approximately 1.24.
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Passage IIA solution containing 10 g of a dibasic acid in 1000 g of water freezes at -0.15° C. 10 mL of the acid is neutralised by 12 mL of 0.1 N NaOH solution. [Kf (H20 ) = 1.86° mol-1kg]Q.van’t Hoff factor (/') of the dibasic acid isa)1.34b)2.00c)2.72d)3.00Correct answer is option 'A'. Can you explain this answer?
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Passage IIA solution containing 10 g of a dibasic acid in 1000 g of water freezes at -0.15° C. 10 mL of the acid is neutralised by 12 mL of 0.1 N NaOH solution. [Kf (H20 ) = 1.86° mol-1kg]Q.van’t Hoff factor (/') of the dibasic acid isa)1.34b)2.00c)2.72d)3.00Correct answer is option 'A'. 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 Passage IIA solution containing 10 g of a dibasic acid in 1000 g of water freezes at -0.15° C. 10 mL of the acid is neutralised by 12 mL of 0.1 N NaOH solution. [Kf (H20 ) = 1.86° mol-1kg]Q.van’t Hoff factor (/') of the dibasic acid isa)1.34b)2.00c)2.72d)3.00Correct answer is option 'A'. 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 Passage IIA solution containing 10 g of a dibasic acid in 1000 g of water freezes at -0.15° C. 10 mL of the acid is neutralised by 12 mL of 0.1 N NaOH solution. [Kf (H20 ) = 1.86° mol-1kg]Q.van’t Hoff factor (/') of the dibasic acid isa)1.34b)2.00c)2.72d)3.00Correct answer is option 'A'. Can you explain this answer?.
Passage IIA solution containing 10 g of a dibasic acid in 1000 g of water freezes at -0.15° C. 10 mL of the acid is neutralised by 12 mL of 0.1 N NaOH solution. [Kf (H20 ) = 1.86° mol-1kg]Q.van’t Hoff factor (/') of the dibasic acid isa)1.34b)2.00c)2.72d)3.00Correct answer is option 'A'. 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 Passage IIA solution containing 10 g of a dibasic acid in 1000 g of water freezes at -0.15° C. 10 mL of the acid is neutralised by 12 mL of 0.1 N NaOH solution. [Kf (H20 ) = 1.86° mol-1kg]Q.van’t Hoff factor (/') of the dibasic acid isa)1.34b)2.00c)2.72d)3.00Correct answer is option 'A'. 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 Passage IIA solution containing 10 g of a dibasic acid in 1000 g of water freezes at -0.15° C. 10 mL of the acid is neutralised by 12 mL of 0.1 N NaOH solution. [Kf (H20 ) = 1.86° mol-1kg]Q.van’t Hoff factor (/') of the dibasic acid isa)1.34b)2.00c)2.72d)3.00Correct answer is option 'A'. Can you explain this answer?.
Solutions for Passage IIA solution containing 10 g of a dibasic acid in 1000 g of water freezes at -0.15° C. 10 mL of the acid is neutralised by 12 mL of 0.1 N NaOH solution. [Kf (H20 ) = 1.86° mol-1kg]Q.van’t Hoff factor (/') of the dibasic acid isa)1.34b)2.00c)2.72d)3.00Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12.
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Passage IIA solution containing 10 g of a dibasic acid in 1000 g of water freezes at -0.15° C. 10 mL of the acid is neutralised by 12 mL of 0.1 N NaOH solution. [Kf (H20 ) = 1.86° mol-1kg]Q.van’t Hoff factor (/') of the dibasic acid isa)1.34b)2.00c)2.72d)3.00Correct answer is option 'A'. Can you explain this answer?, a detailed solution for Passage IIA solution containing 10 g of a dibasic acid in 1000 g of water freezes at -0.15° C. 10 mL of the acid is neutralised by 12 mL of 0.1 N NaOH solution. [Kf (H20 ) = 1.86° mol-1kg]Q.van’t Hoff factor (/') of the dibasic acid isa)1.34b)2.00c)2.72d)3.00Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of Passage IIA solution containing 10 g of a dibasic acid in 1000 g of water freezes at -0.15° C. 10 mL of the acid is neutralised by 12 mL of 0.1 N NaOH solution. [Kf (H20 ) = 1.86° mol-1kg]Q.van’t Hoff factor (/') of the dibasic acid isa)1.34b)2.00c)2.72d)3.00Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
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