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Water saturated with air ( $20 % O_{2}$ and $80 % N_{2}$ ) at $1 a t m$ and 298 K contains $8.9 \times 10^{- 3} {gL}^{- 1}$ dissolved oxygen. Its solubility under $25 a t m$ pressure of $O_{2}$ will be

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8.9 \times 10^{- 3} {gL}^{- 1}

1.1125 {gL}^{- 1}

1.5642 {gL}^{- 1}

1.834 \times 10^{- 2} {gL}^{- 1}

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detailed solution

Correct option is B

Since $p = k_{H} c$ ,

we have $0.20 atm = k_{H} (8.9 \times 10^{- 3} {gL}^{- 1})$ and $25 atm = k_{H} c$

Hence, $c = (8.9 \times 10^{- 3} {gL}^{- 1}) (25 atm) / 0.20 atm = 1.1125 {gL}^{- 1}$

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Water saturated with air (20% O2 and 80% N2) at 1 atm and 298 K contains 8.9×10−3gL−1 dissolved oxygen. Its solubility under 25 atm pressure of O2 will be