Question

Having a 1.77 M aqueous solution of C12H22O11, what is the mole fraction of sucrose at 25°C? Assume that the addition of sucrose did not change the volume of the solution.

Answer 1

In Aqueous solution of sucrose , sucrose is the solute and water is the solvent.

Given molarity of solution is 1.77M , 1.77 moles per litre.

Moles of solute =1.77

Volume of solvent =1 litre= 1000mL

Density of water =1gm/mL

Mass of water = density × volume = 1gm/mL × 1000mL = 1000 gm.

Molar mass of water = 18 gm/mol

Moles = mass ÷ molar mass = 1000gm÷ 18gm/mol = 55.55mol = 55.6 mol

Total moles = Moles of solute + moles of solvent = 1.77 + 55.6 = 57.37Moles

Mole fraction of sucrose = (Moles of sucrose )÷ Total moles = 1.77÷ 57.37 = 0.3085.