Certainly! To calculate the mass of a non-volatile solute required to reduce the vapor pressure of octane, follow these steps:
Understanding Vapor Pressure Reduction
- Vapor pressure reduction occurs when a non-volatile solute is added to a solvent. The solute occupies space and decreases the number of solvent molecules at the surface, thus lowering the vapor pressure.
Given Data
- Molar mass of solute = 40 g/mol
- Mass of octane (solvent) = 114 g
- Desired vapor pressure = 80% of the original vapor pressure.
Calculating Mole Fraction
- The decrease in vapor pressure can be expressed using Raoult's Law.
- If the original vapor pressure of pure octane is P0, then the new vapor pressure (P) is given by:
P = X_solvent * P0
Where X_solvent is the mole fraction of the solvent.
Finding Mole Fraction
- Since P = 0.80 * P0, we get:
0.80 * P0 = X_solvent * P0
Thus, X_solvent = 0.80.
- Therefore, the mole fraction of the solute (X_solute) = 1 - X_solvent = 0.20.
Relating Moles
- Let n_solute be the number of moles of solute and n_solvent be the number of moles of octane.
- Moles of octane (n_solvent) = mass (g) / molar mass = 114 g / 114 g/mol = 1 mol.
- Thus, X_solute = n_solute / (n_solute + n_solvent) = 0.20.
Setting Up the Equation
- 0.20 = n_solute / (n_solute + 1)
- Solving for n_solute gives:
n_solute = 0.20 + 0.20 * n_solute
n_solute (0.80) = 0.20
n_solute = 0.25 mol.
Calculating Mass of Solute
- Mass of solute = n_solute * molar mass = 0.25 mol * 40 g/mol = 10 g.
Conclusion
- To reduce the vapor pressure of 114 g of octane to 80%, 10 g of the non-volatile solute should be dissolved.