What is the evaporated volume of one pint of liquid methanol (MW=46.1, SG=0.79) at standard temperature...

What is the evaporated volume of one pint of liquid methanol (MW=46.1, SG=0.79) at standard temperature and pressure?

Expert Solution

PV = nRT

P=1 atm, V = ? R= 0.082 L atm mol^-1 K^-1 T= 25^oC= 298 K

SG (specific gravity ) = density of the substance/density of pure water

density of the substance= SG * density of pure water

= 0.79 * 1g/ mL

= 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

queen_honey_blossom answered 2 years ago

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