Question

Liquid Halothane has a density of 1.87 g/mL, and boils at 50.2 degrees celsius and 1.00 atm.

A) If Halothane behaved as an ideal gas, what volume would 10.0 mL of Halothane occupy at 60 degrees celsius and 1.00 atm of pressure? (In Liters)

B) What is the density in g/L of Halothane vapor at 55 degrees celsius and 1.00 atm of pressure?

Answer 1

At temperatures >50.2, Halothane is gas. Hence

From Gas law V= nRT/P, n=number of moles , R=0.08206 L.atm/mole.k, T=60deg.c =60+273.15 =333.15 k

P= 1ATM

Mass of halothanew= Volume*density= 10*1.87= 18.7 gm

Molecular weight of halothane= 197.381

Moles =Mass/Molecular weight =18.7/197.381=0.094

VOLUME =0.094*0.08206*333.15/1 L=2.56 L

B) AT 55 deg.c, from gas law PV= nRT= (Mass/ Molecular weight)*R*T

P * Molecular weight = (Mass/Volume)*RT= density*RT

density = P* Molecular weight/ RT= 1*197.381/ (0.08206*(55+273.15) gm/L=7.32 g/L