Question

1.2 You are in a poorly ventilated organic chemistry laboratory. Contrary to laboratory rules, you drink water from an open paper cup in the laboratory! A large bottle of pure toluene has been left open in the lab, outside the fume hood. Assume you have large excess of toluene available in the container. After sufficient time has passed, what is the concentration of toluene in the air that you breathe? Express the concentration in mass/volume. After sufficient time has passed, what is the concentration of toluene in the water present in the paper cup from which you are drinking (assuming the air is saturated with toluene)? Please explain in words on how did the toluene get into your water? (Points 15)

Answer 1

It is given that the mixture contains 10% toluene by mole fraction.

a) Therefore, the mole fraction of toluene in the air = 0.01

b) As per ideal gas equation:

PV = nRT

= (mass of toulene/Mol wt of toluene)*RT

P*Mol wt of toluene/RT = mass/volume

If we assume room temperature conditions

where T = 20C i.e 293K,

then the vapor pressure of toluene = 2.8KPa = 0.028 atm, R = 0.0821 L-atm/mol K; mol wt of toluene = 92 g/mol

mass/volume

= 0.028 * 92/0.0821* 293

= 0.107 g/L