The solubility of toluene (C7H8) is 535 mg/l at 20 C. It has a density of 0.87 g
ID: 820189 • Letter: T
Question
The solubility of toluene (C7H8) is 535 mg/l at 20 C. It has a density of 0.87 g/cm^3 and a Hentry's constant of 0.15 mole/L-atm.
a) One liter of liquid toluene is spilled into 1 m^3 of water. Assuming no loss to the atmosphere (of water or toluene), what is the steady state aqueous concentration of toluene (i.e. the concentration (mg/L) dissolved in the water)?
b) Is there any non-aqueous phase toluene remaining? If so, where will it be located (floating on top of the water, or at the bottom of the tank?)
a) Calculate part a assuming the volume of water is 10m^3.
I am confused because the statement provides henrys constant for toluene and i dont understand how it will be used.
Explanation / Answer
a) Given solubility is 535 mg/L
hence in 1Liter water 535 mg toulene is soluble.
1m^3 = 1000 L
hence amount soluble in 1m^3 = 1000 x 535 mg = 535 gm
moles of toulene = mass/mol mass = 535/92.14 = 5.8
volume of solution = 1000 L
hence conc = moles/volume in L = ( 5.8/1000) = 5.8 x 10^-3 M
b) toulene layers stays above water layer since density of toulene ( 0.87gm/cm^3) is lesser compared to water ( 1gm/cm^3)
c) 10m^3 = 10,000 L
hence mass of toulene dissolved = 5350 gm
moles of toulene 58
Conc = ( 58/10000) = 5.8 x 10^-3
i.e conc doesnot change even if we change volume of water ( however mass of dissolved toulene changes)
Henry law can be applied provided we have sufficient data like partial pressures. However problem can be solved without using it as data given is sufficient.