Search any question & find its solution
Question:
Answered & Verified by Expert
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?
Options:
Solution:
2269 Upvotes
Verified Answer
The correct answer is:
$5.98 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3}$
$\mathrm{K}_{\mathrm{H}}=\frac{\mathrm{S}}{\mathrm{P}} \quad \therefore \mathrm{S}=\mathrm{K}_{\mathrm{H}} \times \mathrm{P}$
$\therefore \mathrm{S}=1.3 \times 10^{-3} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~atm}^{-1} \times 0.46 \mathrm{~atm}=5.98 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{3}$
$\therefore \mathrm{S}=1.3 \times 10^{-3} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~atm}^{-1} \times 0.46 \mathrm{~atm}=5.98 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{3}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.