Search any question & find its solution
Question:
Answered & Verified by Expert
Calculate the pressure of gas if the solubility of gas in water at $25^{\circ} \mathrm{C}$ is $6.85 \times 10^4 \mathrm{~mol} \mathrm{dm}^{-3}$ (Henry's law constant is $6.85 \times 10^4 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{bar}^{-1}$ )
Options:
Solution:
2639 Upvotes
Verified Answer
The correct answer is:
0.853 bar
We know
$$
\mathrm{S}=\mathrm{K}_{\mathrm{H}} \mathrm{P}
$$
In $1 \mathrm{dm}^3$ or 1 lit moles of gas dissolved is $6.85 \times 10^4$ $\therefore \mathrm{p}=6.85 \times 10^4 / 6.85 \times 10^4=1.0$ bar.
$$
\mathrm{S}=\mathrm{K}_{\mathrm{H}} \mathrm{P}
$$
In $1 \mathrm{dm}^3$ or 1 lit moles of gas dissolved is $6.85 \times 10^4$ $\therefore \mathrm{p}=6.85 \times 10^4 / 6.85 \times 10^4=1.0$ bar.
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.