PV = nRT
P=1 atm, V = ?   R= 0.082 L atm mol-1
K-1 T= 25oC= 298 K
SG (specific gravity ) = density of the substance/density of
pure water
density of the substance= SG * density of pure water
                                  Â
= 0.79 * 1g/ mL
                                  Â
= 0.79 g/mL
1 pint = 473.176 mL
mass of the sample = 473.176 mL * 0.79 g/mL = 373.81 g
moles of liquid methanol present = mass/ molar mass = 373.81
g/46.1 = 8.11 mol
PV = nRT
V = nRT/P = 8.11 * 0.082 L atm mol-1 K-1*
298 K/ 1 atm = 198.18 L
volume of evaporated methanol = 198.18 L
