A. A 0.120 m glucose $$ \left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right) $$ solution contains 0.120 mol of glucose in 1.00 kg of water. B. What are the mole fractions of $$ \mathrm{HC}_{3} \mathrm{OH} $$ and H 2 O in the solution. It contains 128 grams of $$ \mathrm{HC}_{3} \mathrm{OH} $$ and 108 grams of $$ \mathrm{H}_{2} \mathrm{O} $$. C. What are the mole fractions of ethanol, $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} $$, and water in a solution prepared by mixing 70.0 g of ethanol with 30.0 g of Water? D. What are the mole fractions of ethanol, C 2 H 5 OH , and water in a solution prepared by mixing 70.0 mL of ethanol with 30.0 mL of water at $$ 25^{\circ} \mathrm{C} $$ ? The density of ethanol is $$ 0.789 \mathrm{~g} / \mathrm{mL} $$ and that of water is $$ 1.00 \mathrm{~g} / \mathrm{mL} $$.

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Below are one set of acceptable answers. In what follows we assume the following approximate molar masses:
 • Glucose, C₆H₁₂O₆: ≈180.16 g/mol (not needed numerically here)
 • Water, H₂O: ≈18.02 g/mol
 • n‐Propanol (here written HC₃OH, with molecular formula C₃H₇OH, i.e. C₃H₈O): ≈60.1 g/mol
 • Ethanol, C₂H₅OH: ≈46.07 g/mol

Note that mole fraction (x) is defined as

xᵢ = (moles of i) / (total moles of all species).

The problems below have four parts:

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Part A. Glucose solution

A 0.120 m solution means 0.120 mol of glucose is dissolved in 1.00 kg (1000 g) of water. (No numerical “question” is asked here, but we can note the numbers.)

The moles of water are
  n(H₂O) = 1000 g ÷ 18.02 g/mol ≈ 55.5 mol.
Thus one has 0.120 (glucose) + 55.5 (water) ≈ 55.62 total moles.

If one wished to give the mole fractions:
  x(glucose) = 0.120/55.62 ≈ 0.00216,
  x(H₂O) = 55.5/55.62 ≈ 0.99784.

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Part B. Mixture of HC₃OH and water

The solution is made by mixing 128 g of HC₃OH (n‐propanol) with 108 g of water.

1. Calculate moles of each:
  n(HC₃OH) = 128 g ÷ 60.1 g/mol ≈ 2.13 mol,
  n(H₂O) = 108 g ÷ 18.02 g/mol ≈ 6.00 mol.

2. Total moles = 2.13 + 6.00 = 8.13 mol.

3. Mole fractions:
  x(HC₃OH) = 2.13/8.13 ≈ 0.262,
  x(H₂O)  = 6.00/8.13 ≈ 0.738.

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Part C. Mixture by mass: 70.0 g ethanol and 30.0 g water

Step 1. Find moles:
  n(ethanol) = 70.0 g ÷ 46.07 g/mol ≈ 1.52 mol,
  n(H₂O)    = 30.0 g ÷ 18.02 g/mol ≈ 1.67 mol.

Step 2. Total moles = 1.52 + 1.67 = 3.19 mol.

Step 3. Mole fractions:
  x(ethanol) = 1.52/3.19 ≈ 0.477,
  x(H₂O)     = 1.67/3.19 ≈ 0.523.

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Part D. Mixture by volume: 70.0 mL ethanol and 30.0 mL water at 25°C

Because the volumes are given, we first convert to mass using the densities.

1. Masses:
  Ethanol: mass = 70.0 mL × 0.789 g/mL = 55.23 g.
  Water: mass = 30.0 mL × 1.00 g/mL = 30.0 g.

2. Moles:
  n(ethanol) = 55.23 g ÷ 46.07 g/mol ≈ 1.20 mol,
  n(H₂O)     = 30.0 g ÷ 18.02 g/mol ≈ 1.67 mol.

3. Total moles = 1.20 + 1.67 = 2.87 mol.

4. Mole fractions:
  x(ethanol) = 1.20/2.87 ≈ 0.418,
  x(H₂O)     = 1.67/2.87 ≈ 0.582.

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Summary of Answers

A. In 1.00 kg water with 0.120 mol glucose:
  x(glucose) ≈ 0.00216  and  x(H₂O) ≈ 0.99784.

B. For 128 g HC₃OH and 108 g water:
  x(HC₃OH) ≈ 0.262  and  x(H₂O) ≈ 0.738.

C. For 70.0 g ethanol and 30.0 g water:
  x(ethanol) ≈ 0.477  and  x(H₂O) ≈ 0.523.

D. For 70.0 mL ethanol (density 0.789 g/mL) and 30.0 mL water:
  x(ethanol) ≈ 0.418  and  x(H₂O) ≈ 0.582.

Any answer that is equivalent within the limits of rounding is correct.
In each part: - **Part A**: 0.00216 mol glucose and 0.99784 mol water. - **Part B**: 0.262 mol n-propanol and 0.738 mol water. - **Part C**: 0.477 mol ethanol and 0.523 mol water. - **Part D**: 0.418 mol ethanol and 0.582 mol water. These mole fractions represent the proportions of each substance in their respective solutions.

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