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A 0.0020 m aqueous solution of an ionic compound Co(NH3)5(NO2)CI freezes at -0.007320C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be :(Kf = + 1.860C/m)
- a)1
- b)2
- c)3
- d)4
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A 0.0020 m aqueous solution of an ionic compound Co(NH3)5(NO2)CI freez...
0.00732 = i x 1.86 x 0.002
∴ i = 1.96 = 2
∴ i = 1.96 = 2
Most Upvoted Answer
A 0.0020 m aqueous solution of an ionic compound Co(NH3)5(NO2)CI freez...
Given:
Molality (m) = 0.0020 m
Freezing point depression (ΔTf) = -0.007320C
Kf = 1.860C/m
To find:
Number of moles of ions produced by 1 mol of Co(NH3)5(NO2)CI
Solution:
We know that for an ionic compound, the number of moles of ions produced is equal to the sum of the moles of cations and anions.
Let x be the number of moles of Co(NH3)5(NO2)CI in the solution.
Molar mass of Co(NH3)5(NO2)CI = (1 x molar mass of Co) + (5 x molar mass of NH3) + (1 x molar mass of NO2) + (1 x molar mass of Cl)
= (1 x 58.93 g/mol) + (5 x 17.03 g/mol) + (1 x 46.01 g/mol) + (1 x 35.45 g/mol)
= 287.23 g/mol
Mass of Co(NH3)5(NO2)CI in 1 L of solution = 0.0020 mol x 287.23 g/mol = 0.574 g
Now, we can use the freezing point depression formula:
ΔTf = Kf x m x i
where i is the van't Hoff factor, which represents the number of particles (ions, molecules, or formula units) produced by 1 mol of the solute.
Rearranging the formula, we get:
i = ΔTf / (Kf x m)
Substituting the given values, we get:
i = -0.007320C / (1.860C/m x 0.0020 m) = 1.97
Since i must be a whole number, we round it to the nearest integer, which is 2.
Therefore, 1 mol of Co(NH3)5(NO2)CI produces 2 moles of ions.
Answer: Option B (2)
Molality (m) = 0.0020 m
Freezing point depression (ΔTf) = -0.007320C
Kf = 1.860C/m
To find:
Number of moles of ions produced by 1 mol of Co(NH3)5(NO2)CI
Solution:
We know that for an ionic compound, the number of moles of ions produced is equal to the sum of the moles of cations and anions.
Let x be the number of moles of Co(NH3)5(NO2)CI in the solution.
Molar mass of Co(NH3)5(NO2)CI = (1 x molar mass of Co) + (5 x molar mass of NH3) + (1 x molar mass of NO2) + (1 x molar mass of Cl)
= (1 x 58.93 g/mol) + (5 x 17.03 g/mol) + (1 x 46.01 g/mol) + (1 x 35.45 g/mol)
= 287.23 g/mol
Mass of Co(NH3)5(NO2)CI in 1 L of solution = 0.0020 mol x 287.23 g/mol = 0.574 g
Now, we can use the freezing point depression formula:
ΔTf = Kf x m x i
where i is the van't Hoff factor, which represents the number of particles (ions, molecules, or formula units) produced by 1 mol of the solute.
Rearranging the formula, we get:
i = ΔTf / (Kf x m)
Substituting the given values, we get:
i = -0.007320C / (1.860C/m x 0.0020 m) = 1.97
Since i must be a whole number, we round it to the nearest integer, which is 2.
Therefore, 1 mol of Co(NH3)5(NO2)CI produces 2 moles of ions.
Answer: Option B (2)
Community Answer
A 0.0020 m aqueous solution of an ionic compound Co(NH3)5(NO2)CI freez...
Given:
molality = 0.002molal
Tf = -0.00732•C
Tf• = 0•C
∆Tf = 0•C-(-0.00732•C)=0.00732•C
∆Tf = i × Kf ×m
i = = ∆Tf/(Kf ×m)
i =0.00732•C/(1.86•C/m × 0.002m)
i =1.9677 ~ 2
molality = 0.002molal
Tf = -0.00732•C
Tf• = 0•C
∆Tf = 0•C-(-0.00732•C)=0.00732•C
∆Tf = i × Kf ×m
i = = ∆Tf/(Kf ×m)
i =0.00732•C/(1.86•C/m × 0.002m)
i =1.9677 ~ 2
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A 0.0020 m aqueous solution of an ionic compound Co(NH3)5(NO2)CI freezes at -0.007320C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be :(Kf = + 1.860C/m) a)1 b)2 c)3 d)4Correct answer is option 'B'. Can you explain this answer?
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A 0.0020 m aqueous solution of an ionic compound Co(NH3)5(NO2)CI freezes at -0.007320C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be :(Kf = + 1.860C/m) a)1 b)2 c)3 d)4Correct 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 A 0.0020 m aqueous solution of an ionic compound Co(NH3)5(NO2)CI freezes at -0.007320C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be :(Kf = + 1.860C/m) a)1 b)2 c)3 d)4Correct 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 A 0.0020 m aqueous solution of an ionic compound Co(NH3)5(NO2)CI freezes at -0.007320C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be :(Kf = + 1.860C/m) a)1 b)2 c)3 d)4Correct answer is option 'B'. Can you explain this answer?.
A 0.0020 m aqueous solution of an ionic compound Co(NH3)5(NO2)CI freezes at -0.007320C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be :(Kf = + 1.860C/m) a)1 b)2 c)3 d)4Correct 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 A 0.0020 m aqueous solution of an ionic compound Co(NH3)5(NO2)CI freezes at -0.007320C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be :(Kf = + 1.860C/m) a)1 b)2 c)3 d)4Correct 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 A 0.0020 m aqueous solution of an ionic compound Co(NH3)5(NO2)CI freezes at -0.007320C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be :(Kf = + 1.860C/m) a)1 b)2 c)3 d)4Correct answer is option 'B'. Can you explain this answer?.
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A 0.0020 m aqueous solution of an ionic compound Co(NH3)5(NO2)CI freezes at -0.007320C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be :(Kf = + 1.860C/m) a)1 b)2 c)3 d)4Correct answer is option 'B'. Can you explain this answer?, a detailed solution for A 0.0020 m aqueous solution of an ionic compound Co(NH3)5(NO2)CI freezes at -0.007320C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be :(Kf = + 1.860C/m) a)1 b)2 c)3 d)4Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of A 0.0020 m aqueous solution of an ionic compound Co(NH3)5(NO2)CI freezes at -0.007320C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be :(Kf = + 1.860C/m) a)1 b)2 c)3 d)4Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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