Question

${{\text{O}}_{\text{2}}}$is bubbled through water at $293\,{\text{K}}$assuming
that${{\text{O}}_{\text{2}}}$exerts a partial pressure of$0.98\,{\text{bar}}$, the solubility of
${{\text{O}}_{\text{2}}}$in ${\text{gm}}{\text{.}}\,{{\text{L}}^{ - 1}}$is (Henry’s law constant$ = \,34\,{\text{k}}\,{\text{bar}}$)
A. $0.025$
B. $0.05$
C. $0.1$
D. $0.2$

Answer 1

Hint: The solubility is determined as the amount of gas dissolved in the volume of the solvent. The amount of the gas can be determined by using Henry’s law and the volume of the solvent can be determined by the density formula.

Step by step answer: According to Henry law, at a constant temperature, the amount of gas dissolved in liquid or mole fraction is directly proportional to the partial pressure of that gas.
We can use the Formula:
 $P = {{\text{K}}_{\text{H}}}\,x$
 ${\text{Density}}\,{\text{ = }}\dfrac{{{\text{Mass}}}}{{{\text{Volume}}}}$
${P_{{O_2}}} \propto {x_{{O_2}}}$
Where,
${P_{{O_2}}}$is the partial pressure of the oxygen gas.
${x_{{O_2}}}$is the mole fraction of oxygen gas.
${P_{{O_2}}} = {{\text{K}}_{\text{H}}}\,{x_{{O_2}}}$

Where,
${{\text{K}}_{\text{H}}}$is Henry’s law constant.
Convert Henry’s law constant from Kbar to bar.
$1\,{\text{k}}\,{\text{bar}}\,{\text{ = }}\,{\text{1000}}\,{\text{bar}}$

Substitute $0.98\,{\text{bar}}$ for partial pressure of oxygen and $34000\,\,{\text{bar}}$for Henry’s law constant.
$0.98\,{\text{bar}} = 34000\,{\text{bar}}\, \times {x_{{O_2}}}$
$\Rightarrow {x_{{O_2}}}\, = \dfrac{{0.98\,{\text{bar}}}}{{34000\,{\text{bar}}}}$
$\Rightarrow {x_{{O_2}}}\, = 2.8 \times {10^{ - 5}}$
So, the mole of oxygen gas is $2.8 \times {10^{ - 5}}$.
Use the mole formula to determine the mass of oxygen gas as follows:
${\text{Mole}}\,\,{\text{ = }}\dfrac{{{\text{Mass}}}}{{{\text{Molar}}\,{\text{mass}}}}$
The molar mass of oxygen gas is $32\,{\text{g/mol}}$.
Substitute $2.8 \times {10^{ - 5}}$ for mole and $32\,{\text{g/mol}}$ for molar mass.
$\Rightarrow 2.8 \times {10^{ - 5}}\,{\text{mol}}\,{\text{ = }}\dfrac{{{\text{Mass}}}}{{32\,{\text{g/mol}}}}$
$\Rightarrow {\text{Mass}}\,{\text{ = }}\,2.8 \times {10^{ - 5}}\,{\text{mol}}\, \times 32\,{\text{g/mol}}$
$\Rightarrow {\text{Mass}}\,{\text{ = }}\,89.9 \times {10^{ - 5}}\,{\text{g}}$
So, the mass of oxygen gas is $89.9 \times {10^{ - 5}}\,{\text{g}}$.

-Determine the mole of water as follows:
The total mole fraction is equal to one so, the mole of water is,
$1\, = \,2.8 \times {10^{ - 5}}\,{\text{mol}}\,{{\text{O}}_{\text{2}}}\, + \,{\text{Mole of water}}$
$\Rightarrow {\text{Mole of water}}\, = \,1 - 2.8 \times {10^{ - 5}}\,{\text{mol}}\,{{\text{O}}_{\text{2}}}\,$
$\Rightarrow {\text{Mole of water}}\, = \,0.99\,\,{\text{mol}}\,$

-Use the mole formula to determine the mass of water as follows:
Molar mass of water is $18\,{\text{g/mol}}$.
Substitute $0.99$ for mole and $18\,{\text{g/mol}}$ for molar mass.
$0.99\,{\text{mol}}\,{\text{ = }}\dfrac{{{\text{Mass}}}}{{18\,{\text{g/mol}}}}$
$\Rightarrow {\text{Mass}}\,{\text{ = }}\,0.99\,{\text{mol}}\, \times 18\,{\text{g/mol}}$
$\Rightarrow {\text{Mass}}\,{\text{ = }}\,17.9\,{\text{g}}$
So, the mass of water is$17.9\,{\text{g}}$.

-Use the density formula to determine the volume of water as follows:
${\text{Density}}\,{\text{ = }}\dfrac{{{\text{Mass}}}}{{{\text{Volume}}}}$
The density of water is $1000\,{\text{g/L}}$.
Substitute $17.9\,{\text{g}}$for the mass of water and $1000\,{\text{g/L}}$for the density of water.
$1000\,{\text{g/L}}\,{\text{ = }}\dfrac{{17.9\,{\text{g}}}}{{{\text{Volume}}}}$
$\Rightarrow {\text{Volume}}\,{\text{ = }}\dfrac{{17.9\,{\text{g}}}}{{1000\,{\text{g/L}}}}$
$\Rightarrow {\text{Volume}}\,\,{\text{ = }}\,\,{\text{0}}{\text{.0179}}\,\,{\text{L}}$
So, the volume of water is ${\text{0}}{\text{.0179}}\,\,{\text{L}}$.
Use the solubility formula to determine the solubility as follows:
${\text{solubility}}\,{\text{ = }}\dfrac{{{\text{gram}}\,{\text{of}}\,{\text{gas}}}}{{{\text{Liter}}\,{\text{of}}\,{\text{solvent}}}}$
Substitute $89.9 \times {10^{ - 5}}\,{\text{g}}$for the mass of oxygen gas and ${\text{0}}{\text{.0179}}\,\,{\text{L}}$for the volume of water.
$\Rightarrow {\text{solubility}}\,{\text{ = }}\dfrac{{89.9 \times {{10}^{ - 5}}\,{\text{g}}}}{{{\text{0}}{\text{.0179}}\,{\text{L}}}}$
$\Rightarrow {\text{solubility}}\,{\text{ = }}\,{\text{0}}{\text{.05}}\,{\text{g/L}}$
So, the solubility of oxygen gas is ${\text{0}}{\text{.05}}\,{\text{g/L}}$.

Therefore, option (B) $0.05$is correct.

Note: Molar mass is the sum of the atomic mass. The mass of water is,
$\left( {1.0\,{\text{g/mol}} \times 2\,{\text{H}}} \right)\, + \left( {16\,{\text{g/mol}} \times 1{\text{O}}} \right)\, = \,18\,{\text{g/mol}}$. The total mole fraction of all the substances of a mixture is equal to one or equal to $100\% $.