The vapor pressure of benzene (C₆H₆,78.12 g/mol)at 25 $\circ$ C is 0.1252 atm.What is the change in the vapor pressure when 10.00 g naphthalene (C₁₀H₈,128.2 g/mol)is dissolved in 0.2000 kg C₆H₆?

Question

Quizplus · Accepted Answer

The change in vapor pressure can be calculated using Raoult's Law. First, calculate the mole fraction of benzene: moles of benzene = 0.2000 kg / 78.12 g/mol = 2.560 moles; moles of naphthalene = 10.00 g / 128.2 g/mol = 0.078 moles. Total moles = 2.560 + 0.078 = 2.638 moles. Mole fraction of benzene = 2.560 / 2.638 = 0.9704. Change in vapor pressure = 0.1252 atm * (1 - 0.9704) = 0.0036 atm.

The Vapor Pressure of Benzene (C6H6,78 ∘\circ∘ C Is 0

The Vapor Pressure of Benzene (C₆H₆,78 $\circ$ C Is 0