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If \mathrm{O}_{2} gas is bubbled through water at \mathrm{303 \mathrm{~K},} the number of millimoles of \mathrm{ \mathrm{O}_{2}} gas that dissolve in \mathrm{ 1\; litre} of water is __________
. (Nearest Integer)

 (Given : Henry's Law constant for \mathrm{\mathrm{O}_{2}} at \mathrm{303 \mathrm{~K}} is \mathrm{46.821 \; k} bar and partial pressure of \mathrm{\mathrm{O}_{2}=0.920} bar )

 (Assume solubility of \mathrm{ \mathrm{O}_{2}} in water is too small, nearly negligible) 

Option: 1

1


Option: 2

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Option: 3

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Option: 4

-


Answers (1)

best_answer

We know the formula
\mathrm{P= K_{H}\times x}

\mathrm{given \: K_H=46.82 \times 10^3 \mathrm{bar}}
             \mathrm{P=0.920\, bar}
             \mathrm{x= mole\, fraction\, of\, O_{2}}
             \mathrm{x=\frac{n_{\mathrm{O}_2}}{n_{\mathrm{O}_2}+n_{H_2 \mathrm{O}}}}
             \mathrm{x=\frac{n_{O_{2}}}{n_{H_2{O}}}}        [ due to solubility  of \mathrm{O_{2}} in water is too small]  

\mathrm{Now, \quad 0.920=46.82 \times 10^3 \times \frac{n_{\mathrm{O}_2}}{n_{\mathrm{H}_2 \mathrm{O}}}}
                \mathrm{0.920 =46.82 \times 10^3 \times \frac{n_{\mathrm{O}_2}}{\frac{1000}{18}}}
                \mathrm{0.920 =46.82 \times \mathrm{nO}_2 \times 18}
                \mathrm{n_{\mathrm{O}_2} =1.09 \times 10^{-3}}
                \mathrm{n_{\mathrm{O}_2} =1.09\, \mathrm{mmol} \approx 1\, \mathrm{mmol}}

Ans = 1
      
              

Posted by

Ritika Jonwal

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