Biphenyl, C12H10, is a nonvolatile, nonionizing solute that is soluble in benzene, C6H6. At 25 °C, the vapor pressure of pure benzene is 100.84 Torr. What is the vapor pressure of a solution made from dissolving 13.7 g of biphenyl in 28.1 g of benzene? Psolution = __ Torr
The Correct Answer and Explanation is:
To calculate the vapor pressure of a solution of biphenyl in benzene, we use Raoult’s Law, which is: Psolution=Xsolvent⋅Psolvent0P_{\text{solution}} = X_{\text{solvent}} \cdot P^0_{\text{solvent}}Psolution=Xsolvent⋅Psolvent0
Where:
- PsolutionP_{\text{solution}}Psolution is the vapor pressure of the solution.
- XsolventX_{\text{solvent}}Xsolvent is the mole fraction of the solvent (benzene).
- Psolvent0P^0_{\text{solvent}}Psolvent0 is the vapor pressure of the pure solvent (benzene), which is given as 100.84 Torr.
Step 1: Molar masses
- Biphenyl (C₁₂H₁₀):
C = 12.01, H = 1.008
Molar mass = (12 × 12.01) + (10 × 1.008) = 144.12 + 10.08 = 154.20 g/mol - Benzene (C₆H₆):
Molar mass = (6 × 12.01) + (6 × 1.008) = 72.06 + 6.048 = 78.11 g/mol
Step 2: Moles of solute and solvent
- Moles of biphenyl = 13.7 g ÷ 154.20 g/mol ≈ 0.0889 mol
- Moles of benzene = 28.1 g ÷ 78.11 g/mol ≈ 0.3596 mol
Step 3: Mole fraction of benzene
Xbenzene=nbenzenenbenzene+nbiphenyl=0.35960.3596+0.0889≈0.35960.4485≈0.8018X_{\text{benzene}} = \frac{n_{\text{benzene}}}{n_{\text{benzene}} + n_{\text{biphenyl}}} = \frac{0.3596}{0.3596 + 0.0889} ≈ \frac{0.3596}{0.4485} ≈ 0.8018Xbenzene=nbenzene+nbiphenylnbenzene=0.3596+0.08890.3596≈0.44850.3596≈0.8018
Step 4: Apply Raoult’s Law
Psolution=Xbenzene⋅Pbenzene0=0.8018⋅100.84≈80.91 TorrP_{\text{solution}} = X_{\text{benzene}} \cdot P^0_{\text{benzene}} = 0.8018 \cdot 100.84 ≈ 80.91 \text{ Torr}Psolution=Xbenzene⋅Pbenzene0=0.8018⋅100.84≈80.91 Torr
✅ Final Answer:
80.91 Torr\boxed{80.91\ \text{Torr}}80.91 Torr
📘 Explanation (300 words)
Raoult’s Law governs how a nonvolatile solute affects the vapor pressure of a solvent. Since biphenyl does not vaporize (nonvolatile) and does not ionize in solution (nonionizing), it only lowers the vapor pressure of benzene by reducing the number of benzene molecules available at the surface to evaporate.
The vapor pressure of a liquid is due to molecules at the surface escaping into the vapor phase. When a solute is dissolved in the solvent, some surface area is occupied by the solute particles, which do not evaporate. As a result, fewer solvent molecules can escape into the gas phase, and the overall vapor pressure decreases.
To quantify this, we calculate the mole fraction of the solvent, which is the ratio of solvent moles to total moles in solution. For this problem, we determined the moles of benzene and biphenyl using their molar masses and then used this to find the mole fraction of benzene.
Multiplying this mole fraction by the vapor pressure of pure benzene gives the new, lower vapor pressure of the solution. This reflects how the presence of biphenyl lowers benzene’s vapor pressure proportionally to how much biphenyl is added.
This is an ideal solution approximation, assuming that the interactions between benzene and biphenyl molecules are similar to those between pure benzene molecules. This approach is common in introductory chemistry to demonstrate colligative properties, which depend only on the number of solute particles, not their identity.
