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For a dilute solution containing 2.5 g of a non–volatile non–electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is 2°C. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take Kb = 0.76 K kg mol-1)
- a)724
- b)740
- c)736
- d)718
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
For a dilute solution containing 2.5 g of a non–volatile non&nda...
B → Solute; A → Solvent
WB = 2.5 g, WA = 100 g
Most Upvoted Answer
For a dilute solution containing 2.5 g of a non–volatile non&nda...
Electrolyte in 500 mL of water, the molarity can be calculated as follows:
1. Convert the mass of the solute to moles. To do this, divide the mass of the solute (2.5 g) by the molar mass of the solute. The molar mass can be found by looking up the atomic masses of the elements in the compound and multiplying them by their respective subscripts in the chemical formula.
2. Determine the volume of the solution in liters. Since the volume is given in milliliters, divide it by 1000 to convert to liters.
3. Calculate the molarity by dividing the moles of solute by the volume of the solution in liters.
Let's assume the non-electrolyte in the solution is glucose (C6H12O6) with a molar mass of 180 g/mol.
1. Moles of solute = 2.5 g / 180 g/mol = 0.0139 mol
2. Volume of solution = 500 mL / 1000 = 0.5 L
3. Molarity = 0.0139 mol / 0.5 L = 0.0278 M
Therefore, the molarity of the dilute solution is 0.0278 M.
1. Convert the mass of the solute to moles. To do this, divide the mass of the solute (2.5 g) by the molar mass of the solute. The molar mass can be found by looking up the atomic masses of the elements in the compound and multiplying them by their respective subscripts in the chemical formula.
2. Determine the volume of the solution in liters. Since the volume is given in milliliters, divide it by 1000 to convert to liters.
3. Calculate the molarity by dividing the moles of solute by the volume of the solution in liters.
Let's assume the non-electrolyte in the solution is glucose (C6H12O6) with a molar mass of 180 g/mol.
1. Moles of solute = 2.5 g / 180 g/mol = 0.0139 mol
2. Volume of solution = 500 mL / 1000 = 0.5 L
3. Molarity = 0.0139 mol / 0.5 L = 0.0278 M
Therefore, the molarity of the dilute solution is 0.0278 M.
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For a dilute solution containing 2.5 g of a non–volatile non–electrolyte solute in 100 g of water, theelevation in boiling point at 1 atm pressure is 2°C. Assuming concentration of solute is much lower than theconcentration of solvent, the vapour pressure (mm of Hg) of the solution is (take Kb = 0.76 K kg mol-1)a)724b)740c)736d)718Correct answer is option 'A'. Can you explain this answer?
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For a dilute solution containing 2.5 g of a non–volatile non–electrolyte solute in 100 g of water, theelevation in boiling point at 1 atm pressure is 2°C. Assuming concentration of solute is much lower than theconcentration of solvent, the vapour pressure (mm of Hg) of the solution is (take Kb = 0.76 K kg mol-1)a)724b)740c)736d)718Correct 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 For a dilute solution containing 2.5 g of a non–volatile non–electrolyte solute in 100 g of water, theelevation in boiling point at 1 atm pressure is 2°C. Assuming concentration of solute is much lower than theconcentration of solvent, the vapour pressure (mm of Hg) of the solution is (take Kb = 0.76 K kg mol-1)a)724b)740c)736d)718Correct 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 For a dilute solution containing 2.5 g of a non–volatile non–electrolyte solute in 100 g of water, theelevation in boiling point at 1 atm pressure is 2°C. Assuming concentration of solute is much lower than theconcentration of solvent, the vapour pressure (mm of Hg) of the solution is (take Kb = 0.76 K kg mol-1)a)724b)740c)736d)718Correct answer is option 'A'. Can you explain this answer?.
For a dilute solution containing 2.5 g of a non–volatile non–electrolyte solute in 100 g of water, theelevation in boiling point at 1 atm pressure is 2°C. Assuming concentration of solute is much lower than theconcentration of solvent, the vapour pressure (mm of Hg) of the solution is (take Kb = 0.76 K kg mol-1)a)724b)740c)736d)718Correct 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 For a dilute solution containing 2.5 g of a non–volatile non–electrolyte solute in 100 g of water, theelevation in boiling point at 1 atm pressure is 2°C. Assuming concentration of solute is much lower than theconcentration of solvent, the vapour pressure (mm of Hg) of the solution is (take Kb = 0.76 K kg mol-1)a)724b)740c)736d)718Correct 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 For a dilute solution containing 2.5 g of a non–volatile non–electrolyte solute in 100 g of water, theelevation in boiling point at 1 atm pressure is 2°C. Assuming concentration of solute is much lower than theconcentration of solvent, the vapour pressure (mm of Hg) of the solution is (take Kb = 0.76 K kg mol-1)a)724b)740c)736d)718Correct answer is option 'A'. Can you explain this answer?.
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