At 20.0°C, the vapor pressure of ethanol is 45.0 torr, and the vapor pressure of methanol is 92.0 torr. What is the vapor pressure at 20.0°C of a solution prepared by mixing 31.0 g methanol and 59.0 g ethanol?
1 Answer
Explanation:
According to Raoult's Law, the vapor pressure of a solution of two volatile components can be calculated by the formula
#P_"total" = chi_A P_A^0 + chi_B P_B^0#
where
#chi_A# and#chi_B# are the mole fractions of the components#P_A^0# and#P_B^0# are the pressures of the pure components
First, calculate the mole fractions of each component.
#"59.0 g ethanol" xx "1 mol" / "46 g ethanol" = "1.28 mol ethanol"#
#"31.0 g methanol" xx "1 mol" / "32 g methanol" = "0.969 mol methanol"#
The solution has
#chi_"ethanol" = "1.28 mol ethanol" / "2.25 mol" = 0.570#
#chi_"methanol" = "0.969 mol methanol" / "2.25 mol" = 0.430#
(We can also say that
Thus, the vapor pressure of the solution is
#P = chi_"ethanol" P_"ethanol"^0 + chi_"methanol" P_"methanol"^0#
#= 0.570 * "45.0 torr" + 0.430 * "92.0 torr"#
# = " 65.2 torr"#