Please answer and include steps to how you go the answer please! How many millil
ID: 1012889 • Letter: P
Question
Please answer and include steps to how you go the answer please!
How many milliliters of Rubbing Alcohol, at 25oC, can be vaporized if 224.26-kJ of heat is applied to a 500-mL beaker containing this rubbing alcohol? (Assume a density of rubbing alcohol is 0.786 g*mL-1 ,molar mass of 60.1 g*mol-1 and Hvap = 39.9 KJ*mol-1 at boiling point & Hvap = 45.4 KJ*mol-1 at 25oC) Make sure to include your units.
The vapor pressure of 103-mL Diethyl ether, in a 500-mL beaker, was measured as a function of temperature, and was found to have a slope of -2,859.19-Kelvin, -constant of 133.3 fraction numerator m L over denominator m o l end fraction , and a R-constant of 0.008314 fraction numerator k J over denominator m o l cross times K end fraction . What is the heat of vaporization, in kJ? Assume Diethyl ether behavior is as an Ideal Gas. Please make sure to include your units..
Explanation / Answer
Volume of rubbing alcohol = 500ml , mass =volume* density = 500*0.786=393 gms
Moles of rubbing alcohol=393/60.1=6.54 moles
LAtent heat of vaporization at 25 deg.c= 45.4 Kj/mol
heat supplied = 224.26 Kj
Moles vaporized = 224.26/45 =4.983 moles
fraction vaporized= 4.983/6.54 =0.76
Volume vaporized= 500*0.76=380 ml
b)
From the plot of ln P Vs 1/T where lnP is vapor pressure and T is absolute temperautre in K.
The slope gives -deltaH/R, where deltaH is latent heat of vaporization
-deltaH/R is given as = -2859.19
deltaH= 2859.19*0.008314=23.77 Kj/mol