We’re being asked to determine the mole fraction of glucose (C_{6}H_{12}O_{6}) 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:*

$\overline{){\mathit{m}}{\mathit{o}}{\mathit{l}}{\mathit{e}}{\mathbf{}}{\mathit{f}}{\mathit{r}}{\mathit{a}}{\mathit{c}}{\mathit{t}}{\mathit{i}}{\mathit{o}}{\mathit{n}}{\mathbf{}}{\mathbf{\left(}}{\mathit{X}}{\mathbf{\right)}}{\mathbf{=}}\frac{\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{e}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{s}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{t}\mathbf{e}}{\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{e}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{s}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}}}$

**We’ll calculate the mole fraction of glucose using the following steps:**

**Step 1**: Dete*rmine 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.

Determine the mole fraction of glucose, C_{6}H_{12}O_{6}, (molecular weight 180.18) in a 1.86 M aqueous glucose solution given the density of the glucose solution is 1.38 g/cm^{3}.

a. 0.0311

b. 0.0237

c. 0.0324

d. 0.0243

e. 0.0480

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