Problem: 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.a. 0.0311b. 0.0237c. 0.0324d. 0.0243e. 0.0480

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FREE Expert Solution

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:

mole fraction (X)=mole of solutemole of 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.

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Problem Details

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.

a. 0.0311

b. 0.0237

c. 0.0324

d. 0.0243

e. 0.0480

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