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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
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
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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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