MHT CET · Chemistry · Solutions
The Henry's law constant for oxygen is \(1 \cdot 3 \times 10^{-3} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~atm}^{-1}\). If partial pressure of oxygen is \(0.46\) atmosphere what is the concentration of dissolved oxygen at \(25^{\circ} \mathrm{C}\) and 1 atm pressure?
- A \(5.98 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3}\)
- B \(3.53 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3}\)
- C \(5.98 \mathrm{~mol} \mathrm{dm}^{-3}\)
- D \(2 \cdot 82 \times 10^{-3} \mathrm{~mol} \mathrm{dm}^{-3}\)
Answer & Solution
Correct Answer
(A) \(5.98 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{K}_{\mathrm{H}}=\frac{\mathrm{S}}{\mathrm{P}} \quad \therefore \mathrm{S}=\mathrm{K}_{\mathrm{H}} \times \mathrm{P}\)
\(\therefore \mathrm{S}=1.3 \times 10^{-3} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~atm}^{-1} \times 0.46 \mathrm{~atm}\) \(=5.98 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{3}\)
\(\therefore \mathrm{S}=1.3 \times 10^{-3} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~atm}^{-1} \times 0.46 \mathrm{~atm}\) \(=5.98 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{3}\)
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