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?

Question

21431 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-10T01:47:41

#"65.2 torr"#

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 #"1.28 mol + 0.969 mol = 2.25 mol"# total, so

#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 #chi_"methanol" = 1 - chi_"ethanol" = 1- 0.570 = 0.430# because mole ratios always sum to #1#.)

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"#