# 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

###### 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:

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. ###### 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

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Mole Fraction concept. You can view video lessons to learn Mole Fraction. Or if you need more Mole Fraction practice, you can also practice Mole Fraction practice problems.

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Our tutors rated the difficulty ofDetermine the mole fraction of glucose, C6H12O6, (molecular ...as medium difficulty.

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Based on our data, we think this problem is relevant for Professor Sotero's class at UCF.