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A current of dry air was passed through a solution of 2.5g of a non volatile substance "X" in 100g of water and then through water alone. The loss of weight of the former was 1.25g and that of the latter was 0.05g . Calculate 1.mole fraction of the solute in solution 2. molecular weight of the solute
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A current of dry air was passed through a solution of 2.5g of a non vo...
Mole Fraction of Solute in Solution
To calculate the mole fraction of the solute in the solution, we need to first find out the mass of the solute dissolved in the water.
Mass of solute = Loss in weight of solution containing solute = 1.25g
Mass of solvent = Loss in weight of water alone = 0.05g
Therefore, the mass of the solution containing the solute = 2.5g
Mass of the solvent = 100g
Total mass of the solution = Mass of solute + Mass of solvent
Total mass of the solution = 2.5g + 100g = 102.5g
Mole fraction of solute = Moles of solute / Total moles of solution
Molecular weight of water = 18g/mol
Moles of solvent = Mass of solvent / Molecular weight of solvent
Moles of solvent = 100g / 18g/mol = 5.56 mol
Moles of solute = Mass of solute / Molecular weight of solute
Molecular weight of solute = (Mass of solute / Moles of solute)
Molecular weight of solute = (1.25g / Moles of solute)
Molecular weight of solute = 1.25g / (Moles of solute / 100g)
Molecular weight of solute = 100g x 1.25g / Moles of solute
Molecular weight of solute = 125g / Moles of solute
Moles of solute = Mass of solute / Molecular weight of solute
Moles of solute = 2.5g / (125g / Moles of solute)
Moles of solute = (2.5g x Moles of solute) / 125g
Moles of solute = 0.02 x Moles of solute
Total moles of solution = Moles of solute + Moles of solvent
Total moles of solution = 0.02 x Moles of solute + 5.56 mol
Mole fraction of solute = Moles of solute / Total moles of solution
Mole fraction of solute = (0.02 x Moles of solute) / (0.02 x Moles of solute + 5.56 mol)
Molecular Weight of Solute
The molecular weight of the solute can be calculated using the formula:
Molecular weight of solute = Mass of solute / Moles of solute
Substitute the values:
Molecular weight of solute = 2.5g / Moles of solute
From the previous calculation, we know that Moles of solute = 0.02 x Moles of solute
Molecular weight of solute = 2.5g / (0.02 x Moles of solute)
Molecular weight of solute = 125g / Moles of solute
Molecular weight of solute = 125g / (0.02 x Moles of solute + 5.56 mol)
Therefore, the mole fraction of the solute in the solution is (0.02 x Moles of solute) / (0.02 x
To calculate the mole fraction of the solute in the solution, we need to first find out the mass of the solute dissolved in the water.
Mass of solute = Loss in weight of solution containing solute = 1.25g
Mass of solvent = Loss in weight of water alone = 0.05g
Therefore, the mass of the solution containing the solute = 2.5g
Mass of the solvent = 100g
Total mass of the solution = Mass of solute + Mass of solvent
Total mass of the solution = 2.5g + 100g = 102.5g
Mole fraction of solute = Moles of solute / Total moles of solution
Molecular weight of water = 18g/mol
Moles of solvent = Mass of solvent / Molecular weight of solvent
Moles of solvent = 100g / 18g/mol = 5.56 mol
Moles of solute = Mass of solute / Molecular weight of solute
Molecular weight of solute = (Mass of solute / Moles of solute)
Molecular weight of solute = (1.25g / Moles of solute)
Molecular weight of solute = 1.25g / (Moles of solute / 100g)
Molecular weight of solute = 100g x 1.25g / Moles of solute
Molecular weight of solute = 125g / Moles of solute
Moles of solute = Mass of solute / Molecular weight of solute
Moles of solute = 2.5g / (125g / Moles of solute)
Moles of solute = (2.5g x Moles of solute) / 125g
Moles of solute = 0.02 x Moles of solute
Total moles of solution = Moles of solute + Moles of solvent
Total moles of solution = 0.02 x Moles of solute + 5.56 mol
Mole fraction of solute = Moles of solute / Total moles of solution
Mole fraction of solute = (0.02 x Moles of solute) / (0.02 x Moles of solute + 5.56 mol)
Molecular Weight of Solute
The molecular weight of the solute can be calculated using the formula:
Molecular weight of solute = Mass of solute / Moles of solute
Substitute the values:
Molecular weight of solute = 2.5g / Moles of solute
From the previous calculation, we know that Moles of solute = 0.02 x Moles of solute
Molecular weight of solute = 2.5g / (0.02 x Moles of solute)
Molecular weight of solute = 125g / Moles of solute
Molecular weight of solute = 125g / (0.02 x Moles of solute + 5.56 mol)
Therefore, the mole fraction of the solute in the solution is (0.02 x Moles of solute) / (0.02 x
Community Answer
A current of dry air was passed through a solution of 2.5g of a non vo...
Partial pressure of pure solvent= 0.005 g, partial pressure of solution = 1.25g
Partial pressure of solvent - partial pressure of solution ÷ partial pressure of solvent = mole fraction of solute
Therefore, relative lowering of vapour pressure= 0.005÷1.255
=0.00398
Moles of water participating = given wt. /molecular mass
= 100/18 =5.55 moles
For dilute solution, Relative lowering of vapour pressure = given wt. Of solute/molecular mass of solute ÷ no. Of moles of solvent
Molecular mass of solute (m) = 2.5÷5.55*0.00398
Molecualr mass of solute = 2.5/0.022 = 113.06 g/ mol
Partial pressure of solvent - partial pressure of solution ÷ partial pressure of solvent = mole fraction of solute
Therefore, relative lowering of vapour pressure= 0.005÷1.255
=0.00398
Moles of water participating = given wt. /molecular mass
= 100/18 =5.55 moles
For dilute solution, Relative lowering of vapour pressure = given wt. Of solute/molecular mass of solute ÷ no. Of moles of solvent
Molecular mass of solute (m) = 2.5÷5.55*0.00398
Molecualr mass of solute = 2.5/0.022 = 113.06 g/ mol
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A current of dry air was passed through a solution of 2.5g of a non volatile substance "X" in 100g of water and then through water alone. The loss of weight of the former was 1.25g and that of the latter was 0.05g . Calculate 1.mole fraction of the solute in solution 2. molecular weight of the solute
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A current of dry air was passed through a solution of 2.5g of a non volatile substance "X" in 100g of water and then through water alone. The loss of weight of the former was 1.25g and that of the latter was 0.05g . Calculate 1.mole fraction of the solute in solution 2. molecular weight of the solute 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 current of dry air was passed through a solution of 2.5g of a non volatile substance "X" in 100g of water and then through water alone. The loss of weight of the former was 1.25g and that of the latter was 0.05g . Calculate 1.mole fraction of the solute in solution 2. molecular weight of the solute covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A current of dry air was passed through a solution of 2.5g of a non volatile substance "X" in 100g of water and then through water alone. The loss of weight of the former was 1.25g and that of the latter was 0.05g . Calculate 1.mole fraction of the solute in solution 2. molecular weight of the solute.
A current of dry air was passed through a solution of 2.5g of a non volatile substance "X" in 100g of water and then through water alone. The loss of weight of the former was 1.25g and that of the latter was 0.05g . Calculate 1.mole fraction of the solute in solution 2. molecular weight of the solute 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 current of dry air was passed through a solution of 2.5g of a non volatile substance "X" in 100g of water and then through water alone. The loss of weight of the former was 1.25g and that of the latter was 0.05g . Calculate 1.mole fraction of the solute in solution 2. molecular weight of the solute covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A current of dry air was passed through a solution of 2.5g of a non volatile substance "X" in 100g of water and then through water alone. The loss of weight of the former was 1.25g and that of the latter was 0.05g . Calculate 1.mole fraction of the solute in solution 2. molecular weight of the solute.
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A current of dry air was passed through a solution of 2.5g of a non volatile substance "X" in 100g of water and then through water alone. The loss of weight of the former was 1.25g and that of the latter was 0.05g . Calculate 1.mole fraction of the solute in solution 2. molecular weight of the solute, a detailed solution for A current of dry air was passed through a solution of 2.5g of a non volatile substance "X" in 100g of water and then through water alone. The loss of weight of the former was 1.25g and that of the latter was 0.05g . Calculate 1.mole fraction of the solute in solution 2. molecular weight of the solute has been provided alongside types of A current of dry air was passed through a solution of 2.5g of a non volatile substance "X" in 100g of water and then through water alone. The loss of weight of the former was 1.25g and that of the latter was 0.05g . Calculate 1.mole fraction of the solute in solution 2. molecular weight of the solute theory, EduRev gives you an
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