A solution is made by mixing 60.00 grams of water with 26.048 g of sodium carbonate. Pirnam, what is the mole fraction of the solvent in this solution? Assume the density of water is 1.0000 g cm-3.


To calculate the mole fraction of the solvent (water) in the solution, we need to follow these steps:

1. **Find the number of moles of water**:
   The molar mass of water (H₂O) is approximately 18.015 g/mol.

   \[
   \text{Moles of water} = \frac{\text{mass of water}}{\text{molar mass of water}} = \frac{60.00 \, \text{g}}{18.015 \, \text{g/mol}} \approx 3.33 \, \text{mol}
   \]

2. **Find the number of moles of sodium carbonate (Na₂CO₃)**:
   The molar mass of sodium carbonate (Na₂CO₃) is approximately 105.99 g/mol.

   \[
   \text{Moles of sodium carbonate} = \frac{\text{mass of sodium carbonate}}{\text{molar mass of sodium carbonate}} = \frac{26.048 \, \text{g}}{105.99 \, \text{g/mol}} \approx 0.246 \, \text{mol}
   \]

3. **Calculate the total number of moles in the solution**:
   \[
   \text{Total moles} = \text{Moles of water} + \text{Moles of sodium carbonate} = 3.33 + 0.246 \approx 3.576 \, \text{mol}
   \]

4. **Calculate the mole fraction of the solvent (water)**:
   The mole fraction of a component is the ratio of the moles of that component to the total number of moles in the solution. For water:

   \[
   \text{Mole fraction of water} = \frac{\text{Moles of water}}{\text{Total moles}} = \frac{3.33}{3.576} \approx 0.931
   \]

Thus, the mole fraction of the solvent (water) in this solution is approximately **0.931**.

Would you like more details or have any questions?

Here are five related questions:
1. How is the mole fraction different from molarity or molality?
2. What is the importance of mole fraction in colligative properties?
3. How would the mole fraction change if the amount of sodium carbonate were doubled?
4. Can the mole fraction ever be greater than 1?
5. How does the density of water affect the calculation of mole fraction?

**Tip:** Mole fraction is a dimensionless quantity and is useful in expressing concentrations when studying colligative properties like vapor pressure lowering and boiling point elevation.

A solution is made by mixing 60.00 grams of water with 26.048 g of sodium carbonate. What is the mole fraction of the solvent in this solution? Assume the density of water is 1.0000 g cm-3.

Learn how to calculate the mole fraction of the solvent in a solution made by mixing 60.00 grams of water with 26.048 g of sodium carbonate. Follow a detailed step-by-step guide, including key concepts like stoichiometry, mole fraction, and relevant formulas. Suitable for students in Grades 10-12.

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