A vessel with a volume of 2 L is halffilled with water and then sealed so it con

Question

A vessel with a volume of 2 L is halffilled with water and then sealed so it
contains 1 L of air and 1 L of water. We will now conduct an experiment to determine
Henry’s constant for toluene, by adding one gram of liquid toluene through a rubber
septum. We then shake the container and wait for equilibrium. A sample of gas is then
extracted from the head space above the water and analyzed using a GC to determine
that the mass concentration of toluene in the gas phase is 97 mg/L. Determine the
Henry’s Law constant in units of M/atm.

Explanation / Answer

Given :

Concentration of toluene = 97 mg/ L

We have to find Henry’s law constant.

The formula is

C = kp

C is solubility , k is Henry’s law constant.

P is partial pressure.

We can rearrange given equation to get Henry’s law constant.

k = C /p

we know have to find it is M/atm so we need concentration in molar

Lets convert conver C in M

M = mol/ L

Mol = mass in g / molar mass = 97 x 10-3 g / 92.14 g per mol = 0.001053 mol

Solubility or molarity of toluene = 0.001053 mol / 1 L = 0.001053 M

Now we have to get pressure.

Pressure is partial pressure toluene in and that is found by using the mole fraction and pure pressure of toluene.

Pressure of pure toluene = 29.1 torr , pressure in atm = 29.1 torr / 760 torr = 0.03829 atm

Mole fraction of toluene : It can be found by using following set up

Mole fraction = 0.001053 mol/ ( 0.001053 mol toluene + mol H2O )

We now can get moles of H2O by using volume and density of the water to get mass of water in g and then by using mass and molar mass we can get moles H2O

Moles of water = 1.0 L x (1000 mL / 1.0L ) x density of water x (1/ molar mass of water )

                        Density of water is approximately = 1.0 g / mL . lets use it.

Moles of water = 1.0 L x (1000 mL / 1.0L ) x (1.0 g/mL )x ( 1 mol H2O / 18.0148 g )

            = 55.5 mol H2O

Lets plug this value to get mol fraction of toluene.

Mole fraction of toluene = 0.001053 mol toluene / ( 0.001053 mol toluene + 55.5 mol H2O )

= 1.896 E-5

Partial pressure of toluene = 1.89 E-5 x Pure pressure = 1.89 E-5 x 0.3829 atm=

Partial pressure of toluene = 7.26 E-6 atm

Lets use this pressure to get the Henry’s law constant

k = 0.001053 / 7.26 E-6 atm

k = 144.97 M / atm

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.