Question

A 1 L glass bottle contains 200 mL of water and 800 mL of air at 25 °C. If 100 µL of pure toluene is injected into the glass vial (which is immediately sealed), what is the concentration of toluene in the air space of the bottle and what is the concentration of toluene in the water? The dimensionless Henry’s Law Constant for toluene at 25 °C is 0.265 (mg/Lair)/mg/Lwater). The specific gravity of toluene liquid (i.e., the density of toluene divided by the density of water) is 0.867.

Answer #1

mass of toulene added = density * volume = 0.867 * 100 *
10^{-6} = 0.0867 g = 867 mg

Let the concentration of toulene in water be x (mg/Lwater) and
that in air be y (mg/Lair).

Then

x * (volume of water) + y * (volume of air) = 867

x * 0.2 + y * 0.8 = 867 ...(1)

Also, y = K_{H}*x

y = 0.265 x ... (2)

Put value of y from equation (2) in equation (1)

0.2x + 0.8*0.265 x = 867

0.412 x = 867

x = 2104.37

y= 557.66

So, concentration of toulene in water = 2104 (mg/Lwater)

& concentration of toulene in air = 558 (mg/Lair)

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