Question

Exposing a 250 mL sample of water at 20.∘C to an atmosphere containing a gaseous solute at 27.59 kPa resulted in the dissolution of 4.51×10−3 g of the solute. Use Henry's law to determine the solubility of this gaseous solute when its pressure is 79.39 kPa.

Answer 1

Answer : The solubility of this gaseous solute will be, $5.18\times 10^{-5}g/mL$

Explanation :

First we have to calculate the concentration of solute.

$\text{Concentration of solute}=\frac{Mass}{Volume}=\frac{4.51\times 10^{-3}g}{250mL}=1.80\times 10^{-5}g/mL$

Now we have to calculate the Henry's law constant.

Using Henry's law :

$C=k_H\times p$

where,

C = concentration of solute = $1.80\times 10^{-5}g/mL$

p = partial pressure = 27.59 kPa

$k_H$ = Henry's law constant = ?

Now put all the given values in the above formula, we get:

$1.80\times 10^{-5}g/mL=k_H\times (27.59kPa)$

$K_H=6.52\times 10^{-7}g/mL.kPa$

Now we have to calculate the solubility of this gaseous solute when its pressure is 79.39 kPa.

$C_{(79.39kPa)}=k_H\times p$

$C_{(79.39kPa)}=(6.52\times 10^{-7}g/mL.kPa)\times (79.39kPa)$

$C_{(79.39kPa)}=5.18\times 10^{-5}g/mL$

Therefore, the solubility of this gaseous solute will be, $5.18\times 10^{-5}g/mL$