Posted by David on .
A 15.0g sample of liquid ethanol, C2H5OH, absorbs 5.13x10^3 J of heat at its normal boiling point, 78.0C. The molar enthalpy of vaporization of ethanol is 39.3kJ/mol. (a) What volume of ethanol vapor is produced? The voume is measured at 78.0C and 1.00atm pressure. (b) What mass of ethanol remians in the liquid state?
Divide 5.13x10^3 J by 39.3kJ/mol. to get the number of moles of ethanol evaporated.
(a) Use the ideal gas low to calculate the volume of that vapor
V = n RT/P
(b) Convert the number of moles vaporized to mass and subtract that from 15.0 g.