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The Mole Fraction of glucose in aqueous solution is 0.2, then molality of solution will be?
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The Mole Fraction of glucose in aqueous solution is 0.2, then molality...
Explanation:

Mole Fraction: It is the ratio of the number of moles of a particular substance to the total number of moles in a solution.

Molality: It is the number of moles of solute per kilogram of solvent.

To calculate the molality of a solution, we need to know the molecular weight of glucose and the mass of the solvent.

The molecular weight of glucose is 180 g/mol.

Calculation:

Let us assume that we have 1000 g of the solution.

The mole fraction of glucose = 0.2

The mole fraction of water = 0.8

Let the number of moles of glucose be x.

Then, the number of moles of water will be (0.8/0.2) x = 4x.

Total number of moles in the solution = x + 4x = 5x.

Molality of the solution = moles of solute / mass of solvent in kg

Mass of glucose = number of moles of glucose x molecular weight of glucose = x x 180 g

Mass of water = 1000 g - x x 180 g = (5x) x 18 g

Molality of the solution = x / (1000 - 180x) kg

Now, we can equate the mole fraction of glucose to the molality of the solution.

Mole fraction of glucose = x / (5x + 4x) = x / 9x = 0.2

Therefore, x = 0.045 moles.

Molality of the solution = 0.045 / (1000 - (0.045 x 180)) kg

Molality of the solution = 0.045 / 991.1 kg

Molality of the solution = 0.0000455 mol/kg

Conclusion:

Therefore, the molality of the solution is 0.0000455 mol/kg.
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The Mole Fraction of glucose in aqueous solution is 0.2, then molality of solution will be?
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