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If Na2SO4 is considered to be completely dissociated into ions in aqueous solution, the change in freezing point of water when 0.01 mole of it is dissolved in 1Kg of water, is? [Kf = 1.86 KKg/mol]
[rounded up to three decimal places]
[rounded up to three decimal places]
Correct answer is between '0.053,0.057'. Can you explain this answer?
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Calculation of Freezing Point Depression:
1. The formula to calculate the freezing point depression is given by:
∆Tf = i * Kf * m
where ∆Tf is the change in freezing point, i is the van't Hoff factor, Kf is the cryoscopic constant, and m is the molality of the solution.
2. For Na2SO4, the van't Hoff factor (i) is 3 because Na2SO4 dissociates into 3 ions in solution (2Na+ and SO4^2-).
3. Given that 0.01 mole of Na2SO4 is dissolved in 1 kg of water:
- The molality (m) of the solution is calculated as:
m = moles of solute / kg of solvent
m = 0.01 / 1 = 0.01 mol/kg
4. Substituting the values in the formula:
∆Tf = 3 * 1.86 * 0.01 = 0.0558
Final Answer:
Therefore, the change in freezing point of water when 0.01 mole of Na2SO4 is dissolved in 1 kg of water is 0.0558 K.
Rounded up to three decimal places, the answer falls between 0.053 and 0.057 K.
1. The formula to calculate the freezing point depression is given by:
∆Tf = i * Kf * m
where ∆Tf is the change in freezing point, i is the van't Hoff factor, Kf is the cryoscopic constant, and m is the molality of the solution.
2. For Na2SO4, the van't Hoff factor (i) is 3 because Na2SO4 dissociates into 3 ions in solution (2Na+ and SO4^2-).
3. Given that 0.01 mole of Na2SO4 is dissolved in 1 kg of water:
- The molality (m) of the solution is calculated as:
m = moles of solute / kg of solvent
m = 0.01 / 1 = 0.01 mol/kg
4. Substituting the values in the formula:
∆Tf = 3 * 1.86 * 0.01 = 0.0558
Final Answer:
Therefore, the change in freezing point of water when 0.01 mole of Na2SO4 is dissolved in 1 kg of water is 0.0558 K.
Rounded up to three decimal places, the answer falls between 0.053 and 0.057 K.
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If Na2SO4 is considered to be completely dissociated into ions in aqueous solution, the change in freezing point of water when 0.01 mole of it is dissolved in 1Kg of water, is? [Kf = 1.86 KKg/mol][rounded up to threedecimal places]Correct answer is between '0.053,0.057'. Can you explain this answer?
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If Na2SO4 is considered to be completely dissociated into ions in aqueous solution, the change in freezing point of water when 0.01 mole of it is dissolved in 1Kg of water, is? [Kf = 1.86 KKg/mol][rounded up to threedecimal places]Correct answer is between '0.053,0.057'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about If Na2SO4 is considered to be completely dissociated into ions in aqueous solution, the change in freezing point of water when 0.01 mole of it is dissolved in 1Kg of water, is? [Kf = 1.86 KKg/mol][rounded up to threedecimal places]Correct answer is between '0.053,0.057'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If Na2SO4 is considered to be completely dissociated into ions in aqueous solution, the change in freezing point of water when 0.01 mole of it is dissolved in 1Kg of water, is? [Kf = 1.86 KKg/mol][rounded up to threedecimal places]Correct answer is between '0.053,0.057'. Can you explain this answer?.
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