Question

Calculate the solubility of benzene in water at 25 c in ppm. the required henry's law constant is 5.6 bar/mol/kg and benzene's saturated vapor pressure is 0.13 bar.

Answer 1

The relationship between pressure and solubility of the gas is given by Henry's law as:

`S_g = kP_g`

where,

`S_g` is the solubility of the gas.

`k` is proportionality constant i.e. Henry's constant.

`P_g` is pressure of the gas.

`k = 5.6 bar/mol/kg` (given)

`P_g = 0.13 bar` (given)

Substituting the values,

`S_g = 5.6 bar/mol/kg* 0.13 bar = 0.728 mole/kg`

To convert
`mole/kg` to
`g/kg`:

Molar mass of benzene,
`C_6H_6` =
`6* 12+6* 1 = 78 g/mol`

`0.728* 78 = 56.784 g/kg`

Now for converting into
`ppm`:

Since,
`1 ppm = 0.001 g/kg`

So,
`56.784* 1000 = 56784 ppm`.

Hence, the solubility of benzene in water at
`25^(o) C` in
`ppm` is
`56784 ppm`.