Chemists often use the Clausius-Clapeyron equation to estimate the vapour pressures and boiling points of pure liquids:
#color(blue)(bar(ul(|color(white)(a/a) ln(P_2/P_1) = (Δ_"vap"H)/R(1/T_1- 1/T_2)color(white)(a/a)|)))" "#
where
#P_1# and #P_2# are the vapour pressures at temperatures
#T_1# and #T_2#
#Δ_"vap"H# = enthalpy of vaporization of the liquid
#R# = Universal Gas Constant
In your problem,
#P_2 = "1.90 atm"#; #P_1 = "1 atm"#
#Δ_"vap"H = "40.66 kJ/mol"# at 100 °C
#R = "0.008 314 kJ·K"^"-1""mol"^"-1"#
#T_2 = "?"#; #T_1 = "100.00 °C" = "373.15 K"#
#ln((1.90 color(red)(cancel(color(black)("atm"))))/(1 color(red)(cancel(color(black)("atm"))))) = (40.66 color(red)(cancel(color(black)("kJ·mol"^"-1"))))/("0.008 314" color(red)(cancel(color(black)("kJ")))·"K"^"-1"color(red)(cancel(color(black)("mol"^"-1"))))(1/("373.15 K") - 1/T_2)#
#0.642 = 13.11 - (4891 "K")/T_2#
#"-12.46" = "-4891 K"/T_2#
#T_2 = "4891 K"/12.46 = "392 K" = "119 °C"#
