Discussion
N2 exerts a partial pressure of 7.648 bar when dissolved in 1 litre of water at 298 K. What is the mole fraction of N2 at same temperature? ( Henry's law constant (KH) for N2 at 298 K = 76.4 k bar)
Answer: C
The mole fraction of nitrogen is \(10^{-4}\).
Henry's law states that the partial pressure of a gas over a solution is proportional to its mole fraction in the solution, \(p = K_{H} x\). Writing both pressures in bar (76.4 kbar = 76400 bar):
\[ \begin{aligned} x &= \frac{p}{K_{H}} = \frac{7.648}{76400} \\ &= 1.0 \times 10^{-4} \end{aligned} \]
The key step is converting kilobar to bar; leaving the constant as 76.4 would give 0.1, which is not even close to a dissolved gas fraction. \(10^{-3}\), \(10^{-5}\) and the other powers are each at least a factor of ten away from the computed value. A large \(K_{H}\) means low solubility, so a small mole fraction is expected for nitrogen.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development