The vapor pressure of benzene (C₆H₆, 78.12 g/mol) at 25 $\degree$C is 0.1252 atm.What is the change in the vapor pressure when 10.00 g of naphthalene (C₁₀H₈, 128.2 g/mol) is dissolved in 0.2000 kg of 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 °\degree° C Is 01252 Atm

The Vapor Pressure of Benzene (C₆H₆, 78 $\degree$ C Is 01252 Atm