Determine the mole fraction of glucose, C6H12O6, (molecular weight 180.18) in a 1.86 M aqueous glucose solution given the density of the glucose solution is 1.38 g/cm3.
We’re being asked to determine the mole fraction of glucose (C6H12O6) in a 1.86 M aqueous glucose solution.
Aqueous means that water is the solvent. Hence, glucose is the solute.
Mole Fraction (X) relates the moles of solute and solvent within a solution:
We’ll calculate the mole fraction of glucose using the following steps:
Step 1: Determine the moles of solute using the molarity of the solution.
Step 2: Calculate the mass of the solvent using the molarity and density of the solution.
Step 3: Calculate the moles of the solvent.
Step 4: Calculate the mole fraction.