We’re being asked to recognize the mole portion of glucose (C6H12O6) in a 1.86 M aqueous glucose solution.

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Aqueous way that water is the solvent. Hence, glucose is the solute.

Mole fraction (X)relates the moles of solute and also solvent within a solution:

mole fraction (X)=mole of solutemole of solution

We’ll calculation the mole portion of glucose utilizing the following steps:

Step 1: Determine the mole of solute using the molarity that the solution.Step 2: calculate the fixed of the solvent using the molarity and density of the solution.Step 3: calculate the mole of the solvent.Step 4: Calculate the mole fraction.

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Problem Details

Determine the mole portion of glucose, C6H12O6, (molecular weight 180.18) in a 1.86 M aqueous glucose solution offered the thickness 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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What scientific principle do you need to know in stimulate to solve this problem?

Our tutors have actually indicated that to settle this difficulty you will require to use the Mole fraction concept. You have the right to view video clip lessons to find out Mole Fraction. Or if girlfriend need much more Mole portion practice, you can also practice Mole portion practice problems.

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

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Our skilled Chemistry tutor, Jules took 5 minutes and also 25 secs to settle this problem. You can follow their measures in the video explanation above.

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Based on ours data, we think this difficulty is appropriate for Professor Sotero's class at UCF.

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