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Henry's Law constant for \mathrm{O}_{2} dissolved in water is 5.28 \times 10^{5} atm at 25^{\circ} \mathrm{C}.If partial pressure of oxygen in air is 0.6 \mathrm{~atm}.What will be the concentration \left ( M \right ) of dissolved oxygen in water in equilibrium with air at 25^{\circ} \mathrm{C} 

Option: 1

\mathrm{M=11.67 \times 10^{-2}}


Option: 2

\mathrm{M=6.13 \times 10^{-5}}


Option: 3

\mathrm{M=9.87 \times 10^{-3}}


Option: 4

\mathrm{M=10.46 \times 10^{-4}}


Answers (1)

\mathrm{K_{\mathrm{H}}=5.28 \times 10^{5}}
\mathrm{P_{\mathrm{O}_{2}}=0.6 \mathrm{~atm}}

\mathrm{P=K_{\mathrm{H} }X}
\mathrm{X_{\mathrm{O}_{2}}=\frac{0.6}{5.28 \times 10^{5}}=11 \times 10^{-7}}

\mathrm{\text { Moles of water }\left(\mathrm{n_{H_2O}}\right)=\frac{1000}{18}=55.8 \mathrm{~mol} }
\mathrm{X_{O_{2}}=\frac{n_{ O_{2}}}{n_{O_{2}}+n_{H_{2} O}} }

Ignoring \mathrm{n_{\mathrm{O}_{2}}} as its very small compared to \mathrm{n_{H_{2}O}}

\mathrm{ X_{\mathrm{O}_{2}}=\frac{n _{\mathrm{O}_{2}}}{n _{\mathrm{H}_{2} \mathrm{O}} }}
\mathrm{ n_{\mathrm{O}_{2}}=X_{O_{2}} \times n _{\mathrm{H}_{2} \mathrm{O} }}
\mathrm{ n_{{\mathrm{O}}_{2}}=11 \times 10^{-7} \times 55.8 }

\mathrm{ M = n_{O_2}/V , V= 1L}
\mathrm{ {\mathrm{M}}=6.13 \times 10^{-5} }

Posted by

Ramraj Saini

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