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5 g of Na2SO4 was dissolved in x g of H2O. The change in freezing point was found to be 3.82oC. If Na2SO4 is 81.5% ionised, the value of x
(Kf for water=1.86oC kg mol−1) is approximate:
(molar mass of S=32 g mol−1 and that of Na=23 g mol−1)
(Kf for water=1.86oC kg mol−1) is approximate:
(molar mass of S=32 g mol−1 and that of Na=23 g mol−1)
- a)15 g
- b)25 g
- c)45 g
- d)65 g
Correct answer is option 'A'. Can you explain this answer?
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5 g of Na2SO4 was dissolved in x g of H2O.The change in freezing point...
-1) is:
We can use the formula:
ΔTf = Kf * molality
where ΔTf is the change in freezing point, Kf is the freezing point depression constant for water, and molality is the molal concentration of the solution (moles of solute per kg of solvent).
First, we need to calculate the molality of the solution:
molality = moles of solute / mass of solvent in kg
We know that 5 g of Na2SO4 was dissolved in x g of H2O, so the mass of solvent is:
mass of solvent = x g - 5 g
To calculate the moles of solute, we need to know the molecular weight of Na2SO4:
Na2SO4 = 2Na + SO4
2Na = 2 x 23 = 46
SO4 = 32 + 4 x 16 = 96
Molecular weight = 46 + 96 = 142 g/mol
So, the moles of Na2SO4 is:
moles of Na2SO4 = 5 g / 142 g/mol = 0.0352 mol
The moles of ions will be:
moles of ions = 0.0352 mol x 2 x 0.815 = 0.0574 mol
(we multiply by 2 because there are 2 ions per formula unit of Na2SO4, and by 0.815 because the salt is 81.5% ionised).
Now we can calculate the molality:
molality = 0.0574 mol / (x g - 5 g) / 1000 g/kg = 0.0574 / (x - 5) kg/mol
Substituting this into the formula for freezing point depression:
3.82 = 1.86 * 0.0574 / (x - 5)
Solving for x:
x = 5.36 g
Therefore, 5 g of Na2SO4 was dissolved in 5.36 g of H2O.
We can use the formula:
ΔTf = Kf * molality
where ΔTf is the change in freezing point, Kf is the freezing point depression constant for water, and molality is the molal concentration of the solution (moles of solute per kg of solvent).
First, we need to calculate the molality of the solution:
molality = moles of solute / mass of solvent in kg
We know that 5 g of Na2SO4 was dissolved in x g of H2O, so the mass of solvent is:
mass of solvent = x g - 5 g
To calculate the moles of solute, we need to know the molecular weight of Na2SO4:
Na2SO4 = 2Na + SO4
2Na = 2 x 23 = 46
SO4 = 32 + 4 x 16 = 96
Molecular weight = 46 + 96 = 142 g/mol
So, the moles of Na2SO4 is:
moles of Na2SO4 = 5 g / 142 g/mol = 0.0352 mol
The moles of ions will be:
moles of ions = 0.0352 mol x 2 x 0.815 = 0.0574 mol
(we multiply by 2 because there are 2 ions per formula unit of Na2SO4, and by 0.815 because the salt is 81.5% ionised).
Now we can calculate the molality:
molality = 0.0574 mol / (x g - 5 g) / 1000 g/kg = 0.0574 / (x - 5) kg/mol
Substituting this into the formula for freezing point depression:
3.82 = 1.86 * 0.0574 / (x - 5)
Solving for x:
x = 5.36 g
Therefore, 5 g of Na2SO4 was dissolved in 5.36 g of H2O.
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5 g of Na2SO4 was dissolved in x g of H2O.The change in freezing point was found to be 3.82oC. If Na2SO4 is 81.5% ionised, the value of x(Kf for water=1.86oC kg mol−1) is approximate:(molar mass of S=32 g mol−1 and that of Na=23 g mol−1)a)15 gb)25 gc)45 gd)65 gCorrect answer is option 'A'. Can you explain this answer?
Question Description
5 g of Na2SO4 was dissolved in x g of H2O.The change in freezing point was found to be 3.82oC. If Na2SO4 is 81.5% ionised, the value of x(Kf for water=1.86oC kg mol−1) is approximate:(molar mass of S=32 g mol−1 and that of Na=23 g mol−1)a)15 gb)25 gc)45 gd)65 gCorrect 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 5 g of Na2SO4 was dissolved in x g of H2O.The change in freezing point was found to be 3.82oC. If Na2SO4 is 81.5% ionised, the value of x(Kf for water=1.86oC kg mol−1) is approximate:(molar mass of S=32 g mol−1 and that of Na=23 g mol−1)a)15 gb)25 gc)45 gd)65 gCorrect 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 5 g of Na2SO4 was dissolved in x g of H2O.The change in freezing point was found to be 3.82oC. If Na2SO4 is 81.5% ionised, the value of x(Kf for water=1.86oC kg mol−1) is approximate:(molar mass of S=32 g mol−1 and that of Na=23 g mol−1)a)15 gb)25 gc)45 gd)65 gCorrect answer is option 'A'. Can you explain this answer?.
5 g of Na2SO4 was dissolved in x g of H2O.The change in freezing point was found to be 3.82oC. If Na2SO4 is 81.5% ionised, the value of x(Kf for water=1.86oC kg mol−1) is approximate:(molar mass of S=32 g mol−1 and that of Na=23 g mol−1)a)15 gb)25 gc)45 gd)65 gCorrect 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 5 g of Na2SO4 was dissolved in x g of H2O.The change in freezing point was found to be 3.82oC. If Na2SO4 is 81.5% ionised, the value of x(Kf for water=1.86oC kg mol−1) is approximate:(molar mass of S=32 g mol−1 and that of Na=23 g mol−1)a)15 gb)25 gc)45 gd)65 gCorrect 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 5 g of Na2SO4 was dissolved in x g of H2O.The change in freezing point was found to be 3.82oC. If Na2SO4 is 81.5% ionised, the value of x(Kf for water=1.86oC kg mol−1) is approximate:(molar mass of S=32 g mol−1 and that of Na=23 g mol−1)a)15 gb)25 gc)45 gd)65 gCorrect answer is option 'A'. Can you explain this answer?.
Solutions for 5 g of Na2SO4 was dissolved in x g of H2O.The change in freezing point was found to be 3.82oC. If Na2SO4 is 81.5% ionised, the value of x(Kf for water=1.86oC kg mol−1) is approximate:(molar mass of S=32 g mol−1 and that of Na=23 g mol−1)a)15 gb)25 gc)45 gd)65 gCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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ample number of questions to practice 5 g of Na2SO4 was dissolved in x g of H2O.The change in freezing point was found to be 3.82oC. If Na2SO4 is 81.5% ionised, the value of x(Kf for water=1.86oC kg mol−1) is approximate:(molar mass of S=32 g mol−1 and that of Na=23 g mol−1)a)15 gb)25 gc)45 gd)65 gCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.
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