Question:
In an aqueous glucose solution, the mole fraction of water is 40 times the mole fraction of glucose. What is the weight percentage (w/w) of glucose in the solution?
In an aqueous glucose solution, the mole fraction of water is 40 times the mole fraction of glucose. What is the weight percentage (w/w) of glucose in the solution?
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When dealing with mole fractions and weight percentages, relate the moles of each component to their masses using molar masses. Use the given ratio of mole fractions to simplify calculations.
Updated On: Jun 5, 2025
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The Correct Option is C
Solution and Explanation
Step 1: Known Information.
The mole fraction of water (\( X_{\text{water}} \)) is 40 times the mole fraction of glucose (\( X_{\text{glucose}} \)): $$ X_{\text{water}} = 40 \cdot X_{\text{glucose}} $$ Mole fraction is defined as: $$ X_{\text{component}} = \frac{\text{moles of component}}{\text{total moles of all components}} $$ Step 2: Express Mole Fractions.
Let:
\( n_{\text{water}} \): Moles of water
\( n_{\text{glucose}} \): Moles of glucose
The total number of moles in the solution is: $$ n_{\text{total}} = n_{\text{water}} + n_{\text{glucose}} $$ The mole fractions are: $$ X_{\text{water}} = \frac{n_{\text{water}}}{n_{\text{total}}}, \quad X_{\text{glucose}} = \frac{n_{\text{glucose}}}{n_{\text{total}}} $$ Given: $$ X_{\text{water}} = 40 \cdot X_{\text{glucose}} $$ Substitute the expressions for mole fractions: $$ \frac{n_{\text{water}}}{n_{\text{total}}} = 40 \cdot \frac{n_{\text{glucose}}}{n_{\text{total}}} $$ Simplify: $$ n_{\text{water}} = 40 \cdot n_{\text{glucose}} $$ Step 3: Calculate the Weight Percentage (w/w) of Glucose.
The weight percentage (w/w) of glucose is given by:
$$ \text{Weight \% of glucose} = \left( \frac{\text{mass of glucose}}{\text{total mass of solution}} \right) \times 100 $$ Step 3.1: Relate Moles to Masses.
Molar mass of water (\( M_{\text{water}} \)): \( 18 \, \text{g/mol} \)
Molar mass of glucose (\( M_{\text{glucose}} \)): \( 180 \, \text{g/mol} \)
Mass of water: $$ \text{Mass of water} = n_{\text{water}} \cdot M_{\text{water}} = n_{\text{water}} \cdot 18 $$ Mass of glucose: $$ \text{Mass of glucose} = n_{\text{glucose}} \cdot M_{\text{glucose}} = n_{\text{glucose}} \cdot 180 $$ Total mass of the solution: $$ \text{Total mass} = \text{Mass of water} + \text{Mass of glucose} = n_{\text{water}} \cdot 18 + n_{\text{glucose}} \cdot 180 $$ Step 3.2: Substitute \( n_{\text{water}} = 40 \cdot n_{\text{glucose}} \).
From Step 2, we know: $$ n_{\text{water}} = 40 \cdot n_{\text{glucose}} $$ Substitute this into the expressions for mass: $$ \text{Mass of water} = 40 \cdot n_{\text{glucose}} \cdot 18 = 720 \cdot n_{\text{glucose}} $$ $$ \text{Mass of glucose} = n_{\text{glucose}} \cdot 180 $$ $$ \text{Total mass} = 720 \cdot n_{\text{glucose}} + 180 \cdot n_{\text{glucose}} = 900 \cdot n_{\text{glucose}} $$ Step 3.3: Calculate the Weight Percentage.
The weight percentage of glucose is: $$ \text{Weight \% of glucose} = \left( \frac{\text{Mass of glucose}}{\text{Total mass}} \right) \times 100 $$ Substitute the values: $$ \text{Weight \% of glucose} = \left( \frac{180 \cdot n_{\text{glucose}}}{900 \cdot n_{\text{glucose}}} \right) \times 100 $$ Simplify: $$ \text{Weight \% of glucose} = \left( \frac{180}{900} \right) \times 100 = \frac{180}{9} = 20 $$ Final Answer: \( \boxed{20} \)
The mole fraction of water (\( X_{\text{water}} \)) is 40 times the mole fraction of glucose (\( X_{\text{glucose}} \)): $$ X_{\text{water}} = 40 \cdot X_{\text{glucose}} $$ Mole fraction is defined as: $$ X_{\text{component}} = \frac{\text{moles of component}}{\text{total moles of all components}} $$ Step 2: Express Mole Fractions.
Let:
\( n_{\text{water}} \): Moles of water
\( n_{\text{glucose}} \): Moles of glucose
The total number of moles in the solution is: $$ n_{\text{total}} = n_{\text{water}} + n_{\text{glucose}} $$ The mole fractions are: $$ X_{\text{water}} = \frac{n_{\text{water}}}{n_{\text{total}}}, \quad X_{\text{glucose}} = \frac{n_{\text{glucose}}}{n_{\text{total}}} $$ Given: $$ X_{\text{water}} = 40 \cdot X_{\text{glucose}} $$ Substitute the expressions for mole fractions: $$ \frac{n_{\text{water}}}{n_{\text{total}}} = 40 \cdot \frac{n_{\text{glucose}}}{n_{\text{total}}} $$ Simplify: $$ n_{\text{water}} = 40 \cdot n_{\text{glucose}} $$ Step 3: Calculate the Weight Percentage (w/w) of Glucose.
The weight percentage (w/w) of glucose is given by:
$$ \text{Weight \% of glucose} = \left( \frac{\text{mass of glucose}}{\text{total mass of solution}} \right) \times 100 $$ Step 3.1: Relate Moles to Masses.
Molar mass of water (\( M_{\text{water}} \)): \( 18 \, \text{g/mol} \)
Molar mass of glucose (\( M_{\text{glucose}} \)): \( 180 \, \text{g/mol} \)
Mass of water: $$ \text{Mass of water} = n_{\text{water}} \cdot M_{\text{water}} = n_{\text{water}} \cdot 18 $$ Mass of glucose: $$ \text{Mass of glucose} = n_{\text{glucose}} \cdot M_{\text{glucose}} = n_{\text{glucose}} \cdot 180 $$ Total mass of the solution: $$ \text{Total mass} = \text{Mass of water} + \text{Mass of glucose} = n_{\text{water}} \cdot 18 + n_{\text{glucose}} \cdot 180 $$ Step 3.2: Substitute \( n_{\text{water}} = 40 \cdot n_{\text{glucose}} \).
From Step 2, we know: $$ n_{\text{water}} = 40 \cdot n_{\text{glucose}} $$ Substitute this into the expressions for mass: $$ \text{Mass of water} = 40 \cdot n_{\text{glucose}} \cdot 18 = 720 \cdot n_{\text{glucose}} $$ $$ \text{Mass of glucose} = n_{\text{glucose}} \cdot 180 $$ $$ \text{Total mass} = 720 \cdot n_{\text{glucose}} + 180 \cdot n_{\text{glucose}} = 900 \cdot n_{\text{glucose}} $$ Step 3.3: Calculate the Weight Percentage.
The weight percentage of glucose is: $$ \text{Weight \% of glucose} = \left( \frac{\text{Mass of glucose}}{\text{Total mass}} \right) \times 100 $$ Substitute the values: $$ \text{Weight \% of glucose} = \left( \frac{180 \cdot n_{\text{glucose}}}{900 \cdot n_{\text{glucose}}} \right) \times 100 $$ Simplify: $$ \text{Weight \% of glucose} = \left( \frac{180}{900} \right) \times 100 = \frac{180}{9} = 20 $$ Final Answer: \( \boxed{20} \)
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