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,

$S_g = 5.6 bar/mol/kg\times 0.13 bar = 0.728 mole/kg$

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

Molar mass of benzene, $C_6H_6$ = $6\times 12+6\times 1 = 78 g/mol$

$0.728\times 78 = 56.784 g/kg$

Now for converting into $ppm$:

Since, $1 ppm = 0.001 g/kg$

So, $56.784\times 1000 = 56784 ppm$.

Hence, the solubility of benzene in water at $25^{o} C$ in $ppm$ is $56784 ppm$.