Question

Liquid benzene (C6H6) has a density of 0.8765 g/mL at 40 °C. You inject 10.0 mL of liquid benzene into an empty, rigid 7.8 L vess el at 40 °C. You connect the vessel to a mercury manometer and notice that the mercury is at the same level on both sides (see figure labelled “Before”). The next day, you come back to check the vessel, which is still at 40 °C. You notice that some of the benzene has evaporated and that the mercury level has shifted by 183 mm Hg (see figure labelled “After”). How much benzene (in moles) remains in the liquid phase?You may wish to assume that the barometric pressure has stayed constant

After: Before: Pbar P gas Pbar Gas Gas AP P gas

Answer 1

Hi
We can assume that the pressure difference is caused because of give evaporation of the
Benzene.
Pressure generated because of vaporization = 183 mm Hg
=183/170 atm=0.24 atm
= 0.2*101325
=24397 Pa
=PV=NRT
N=PV/RT
= (24397*7.8*0.001)/(8.314*313)
=0.073 moles
Mass = moles*molecular mass
=0.073*78
=5.7 grams

Initial amount of benzene = volume*density
=10*0.875 = 8.7 g
Moles remaining in the liquid phase
=8.7-5.7= 3 grams

Hope this helps
Answer is subject to the approximation taken in the steps while calculation.
Thanks