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A 0. 00 20 m aqueous solution of an ionic compound Co(NH3)5(NO2)Cl freezes at – 0.00732 °C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be (Kf = – 1.86°C/m)
- a)3
- b)4 [2009]
- c)1
- d)2
Correct answer is option 'D'. Can you explain this answer?
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A 0. 00 20 m aqueous solution of an ionic compound Co(NH3)5(NO2)Cl fre...
ΔTf = 0 – (0.00732°) = 0.00732
ΔTf = i × Kf × m
ΔTf = i × Kf × m
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A 0. 00 20 m aqueous solution of an ionic compound Co(NH3)5(NO2)Cl fre...
The freezing point of a solution is determined by the concentration of solute particles in the solution. In this case, the solute particles are the ions from the ionic compound Co(NH3)5(NO2)Cl.
To calculate the freezing point of the solution, we need to know the van't Hoff factor (i), which represents the number of particles that a solute molecule dissociates into in solution.
In this case, Co(NH3)5(NO2)Cl dissociates into Co2+, 5 NH3, NO2-, and Cl-. So the van't Hoff factor (i) is 7.
The freezing point depression (ΔTf) can be calculated using the formula:
ΔTf = i * Kf * m
Where:
ΔTf = freezing point depression
i = van't Hoff factor
Kf = cryoscopic constant of the solvent
m = molality of the solution
Since the solution is aqueous, the Kf value for water is used. The Kf value for water is 1.86 °C/m.
Given that the molality (m) of the solution is 0.020 m, we can calculate the freezing point depression:
ΔTf = 7 * 1.86 °C/m * 0.020 m
ΔTf = 0.2604 °C
Therefore, the freezing point of the 0.020 m aqueous solution of Co(NH3)5(NO2)Cl would be lowered by 0.2604 °C.
To calculate the freezing point of the solution, we need to know the van't Hoff factor (i), which represents the number of particles that a solute molecule dissociates into in solution.
In this case, Co(NH3)5(NO2)Cl dissociates into Co2+, 5 NH3, NO2-, and Cl-. So the van't Hoff factor (i) is 7.
The freezing point depression (ΔTf) can be calculated using the formula:
ΔTf = i * Kf * m
Where:
ΔTf = freezing point depression
i = van't Hoff factor
Kf = cryoscopic constant of the solvent
m = molality of the solution
Since the solution is aqueous, the Kf value for water is used. The Kf value for water is 1.86 °C/m.
Given that the molality (m) of the solution is 0.020 m, we can calculate the freezing point depression:
ΔTf = 7 * 1.86 °C/m * 0.020 m
ΔTf = 0.2604 °C
Therefore, the freezing point of the 0.020 m aqueous solution of Co(NH3)5(NO2)Cl would be lowered by 0.2604 °C.
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A 0. 00 20 m aqueous solution of an ionic compound Co(NH3)5(NO2)Cl freezes at – 0.00732 °C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be (Kf = – 1.86°C/m)a)3b)4 [2009]c)1d)2Correct answer is option 'D'. Can you explain this answer?
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A 0. 00 20 m aqueous solution of an ionic compound Co(NH3)5(NO2)Cl freezes at – 0.00732 °C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be (Kf = – 1.86°C/m)a)3b)4 [2009]c)1d)2Correct answer is option 'D'. 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 A 0. 00 20 m aqueous solution of an ionic compound Co(NH3)5(NO2)Cl freezes at – 0.00732 °C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be (Kf = – 1.86°C/m)a)3b)4 [2009]c)1d)2Correct answer is option 'D'. 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 A 0. 00 20 m aqueous solution of an ionic compound Co(NH3)5(NO2)Cl freezes at – 0.00732 °C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be (Kf = – 1.86°C/m)a)3b)4 [2009]c)1d)2Correct answer is option 'D'. Can you explain this answer?.
A 0. 00 20 m aqueous solution of an ionic compound Co(NH3)5(NO2)Cl freezes at – 0.00732 °C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be (Kf = – 1.86°C/m)a)3b)4 [2009]c)1d)2Correct answer is option 'D'. 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 A 0. 00 20 m aqueous solution of an ionic compound Co(NH3)5(NO2)Cl freezes at – 0.00732 °C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be (Kf = – 1.86°C/m)a)3b)4 [2009]c)1d)2Correct answer is option 'D'. 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 A 0. 00 20 m aqueous solution of an ionic compound Co(NH3)5(NO2)Cl freezes at – 0.00732 °C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be (Kf = – 1.86°C/m)a)3b)4 [2009]c)1d)2Correct answer is option 'D'. Can you explain this answer?.
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