Question

At what partial pressure oxygen will have a solubility of 0.05 gram per litre in water at 293 kelvin. Henry's constant kh in h2o at 293 kelvin is 34.86. Assume density of solution is same as that of solvent.

Answer 1

Correct question :

At what partial pressure oxygen will have a solubility of $\rm{0.05\:g\:L^{-1}}$ in water at 293 K? Henry's constant ($\rm{K_H}$) for $\rm{O_2}$ in water at 293 K is 34.86 kbar. Assume density of solution is same as that of solvent.

Answer :

• Partial pressure = 0.98 bar.

Step-by-step explanation :

Given that,

• Henry's constant for $\bold{O_2}$ in water (at 293 K) = 34.86 kbar = $\bold{34.86\times{}10^3\:bar}$

$\hrulefill$

✒ Calculation of mole fraction ($\bold{x_{O_2}}$) :

Mass of 1 L of solution = 1000 g.

∴ Mass of solvent (water) =

$\bold{1000\:g-0.05\:g\simeq1000\:g}$

∴ $\bold{n_{H_2O}=\dfrac{1000\:g}{18\:g\:mol^{-1}}=55.5\:moles}$

Now, $\bold{n_{O_2}=\dfrac{0.05\:g}{32\:g\:mol^{-1}}=1.56\times{}10^{-3}\:mole}$

∴ $\bold{x_{O_2}=\dfrac{n_{O_2}}{n_{O_2}+n_{H_2O}}\simeq\dfrac{n_{O_2}}{n_{H_2O}}}$

$\implies\bold{x_{O_2}=\dfrac{1.56\times{}10^{-3}}{55.5}}$

$\implies\bold{x_{O_2}=2.81\times{}10^{-5}}$

$\rule{200}{3}$

✒ Calculation of partial pressure $\bold{(p_{O_2})}$ :

Now, we have to find partial pressure. Applying Henry's law,

$\bold{p_{O_2}=K_H\times{}x_{O_2}}$

$\implies\bold{p_{O_2}=(34.86\times{}10^3\:bar)\times{}(2.81\times{}10^{-5})}$

$\implies\bold{p_{O_2}=0.98\:bar.}$