Calculation of Molarity, Molality, %w/w, and %w/v of Aqueous Glucose Solution

Given:
Mole fraction of glucose solution = 0.1
Specific gravity of solution = 1.1

(i) Calculation of Molarity:
Mole fraction of glucose solution = 0.1
It means, mole fraction of water = 0.9 (as mole fraction of a solution is equal to the mole fraction of solute and solvent)
Thus, total moles of solution = moles of glucose + moles of water
Let's assume the total mass of the solution to be 1 kg
Mass of glucose in solution = mole fraction of glucose x total mass of solution x molar mass of glucose
= 0.1 x 1000 g x 180 g/mol (molar mass of glucose)
= 18000 g = 18 kg
Moles of glucose in solution = mass of glucose/molar mass of glucose
= 18 kg/180 g/mol = 100 moles
Moles of water in solution = total moles of solution - moles of glucose
= 1000 g/18 g/mol - 100 = 44.44 moles
Molarity of glucose solution = moles of glucose/volume of solution in liters
Let's assume the volume of the solution to be 1 L
Molarity of glucose solution = 100 mol/1 L = 100 M

(ii) Calculation of Molality:
Molality of glucose solution = moles of glucose/mass of solvent in kg
Mass of solvent in solution = total mass of solution - mass of glucose in solution
= 1 kg - 18 kg = 0.982 kg
Molality of glucose solution = 100 mol/0.982 kg = 101.82 mol/kg

(iii) Calculation of %w/w:
%w/w of glucose in solution = mass of glucose/mass of solution x 100%
%w/w of glucose in solution = 18 kg/1 kg x 100% = 1800%

(iv) Calculation of %w/v:
%w/v of glucose in solution = mass of glucose/volume of solution x 100%
%w/v of glucose in solution = 18 kg/1 L x 100% = 1800%

Conclusion:
For the given aqueous glucose solution with a mole fraction of 0.1 and a specific gravity of 1.1, molarity is 100 M, molality is 101.82 mol/kg, %w/w is 1800%, and %w/v is 1800%.