Question:
The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of \(20\%\) is to \(79\%\) by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if the Henry’s law constants for oxygen and nitrogen at 298 K are \(3.30 × 107 mm\) and \(6.51 × 107 mm\) respectively, calculate the composition of these gases in water
The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of \(20\%\) is to \(79\%\) by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if the Henry’s law constants for oxygen and nitrogen at 298 K are \(3.30 × 107 mm\) and \(6.51 × 107 mm\) respectively, calculate the composition of these gases in water
Updated On: Sep 29, 2023
Hide Solution
Verified By Collegedunia
Solution and Explanation
Percentage of oxygen \((O_2)\) in air = \(20 \%\)
Percentage of nitrogen \((N_2)\) in air = \(79\%\)
Also, it is given that water is in equilibrium with air at a total pressure of 10 atm, that is, \((10 × 760) mm\, Hg = 7600 mm\, Hg \)
Therefore,
Partial pressure of oxygen, \(p^o_2=\frac{20}{100} \times 7600mmHg\)
= 1520 mm Hg
Partial pressure of nitrogen, \(p_{N_2}=\frac{79}{100} \times 7600mmHg\)
= 6004 mmHg
Now, according to Henry's law:
\(p = K_H.x \)
For oxygen:
\(p^o_2=K_H.x_{O_2}\)
\(⇒x_{O_2}=\frac{p_{O_2}}{K_H}\)
\(=\frac{1520mmHg}{3.30 \times 10^7mmHg}\) (Given \(K_H=3.30 \times 10^7mmHg\))
\(=461 \times 10^{-5}\)
For nitrogen:
\(p_{N_2}=K_H.x_{N_2}\)
\(⇒x_{N_2}=\frac{p_{N_2}}{K_H}\)
\(=\frac{6004mm\,Hg}{6.51 \times 107mm\,Hg}\)
\(=9.22 \times 10^{-5}\)
Hence, the mole fractions of oxygen and nitrogen in water are \(4.61 ×10^{-5}\) and \(9.22 × 10^{-5}\) respectively.
Percentage of nitrogen \((N_2)\) in air = \(79\%\)
Also, it is given that water is in equilibrium with air at a total pressure of 10 atm, that is, \((10 × 760) mm\, Hg = 7600 mm\, Hg \)
Therefore,
Partial pressure of oxygen, \(p^o_2=\frac{20}{100} \times 7600mmHg\)
= 1520 mm Hg
Partial pressure of nitrogen, \(p_{N_2}=\frac{79}{100} \times 7600mmHg\)
= 6004 mmHg
Now, according to Henry's law:
\(p = K_H.x \)
For oxygen:
\(p^o_2=K_H.x_{O_2}\)
\(⇒x_{O_2}=\frac{p_{O_2}}{K_H}\)
\(=\frac{1520mmHg}{3.30 \times 10^7mmHg}\) (Given \(K_H=3.30 \times 10^7mmHg\))
\(=461 \times 10^{-5}\)
For nitrogen:
\(p_{N_2}=K_H.x_{N_2}\)
\(⇒x_{N_2}=\frac{p_{N_2}}{K_H}\)
\(=\frac{6004mm\,Hg}{6.51 \times 107mm\,Hg}\)
\(=9.22 \times 10^{-5}\)
Hence, the mole fractions of oxygen and nitrogen in water are \(4.61 ×10^{-5}\) and \(9.22 × 10^{-5}\) respectively.
Was this answer helpful?
0
0
Top Questions on Solutions
- Density of 3 M NaCl solution is 1.25 g/mL. The molality of the solution is:
- 20 mL of 2 M NaOH solution is added to 400 mL of 0.5 M NaOH solution. The final concentration of the solution is \(\_\_\_\_\_\) x \(10^{-2}\)M. (Nearest integer)
- A solution is prepared by dissolving 5 g of a non-volatile solute in 200 g of water. It has a vapor pressure of 31.84 mm Hg at 300 K. Calculate the molar mass of the solute.
(Vapor pressure of pure water at 300 K = 32 mm Hg) - What type of deviation is shown by ethanol and acetone mixture ? Give reason. What type of azeotropic mixture is formed by that deviation ?
- Define molal depression constant. How is it related to enthalpy of fusion?
View More Questions
Questions Asked in CBSE CLASS XII exam
- Which fruit has the maximum vitamin A content?
- CBSE CLASS XII - 2024
- Nutrition
- Write the reactions of glucose with
(a) HI
(b)\((CH_3CO_2)O\)- CBSE CLASS XII - 2024
- Biomolecules
- Ankur Sachdeva did his MBA from ITB University. He decided to apply his knowledge of scientific management in the fast food restaurant chain 'Coffee Bean' set up by him. This restaurant was providing burgers, fries, shakes etc, as a part of its menu. Nowadays people are quality conscious, so he was using standardised raw materials, processes, methods, working conditions, machinery etc. The objective was to establish standards of excellence. By doing this he was not only able to reduce the cost but was also able to provide new varieties of burgers, fries and shakes leading to increased turnover. Ankur Sachdeva also believed that there was only one best method to maximise efficiency. As a result he developed best way of grilling burgers, cooking fries and preparing shakes. His main objective was to maximise the satisfaction of customers, which he was able to achieve. Not only to learn the best way of doing a job, but to perform their tasks efficiently, 'Coffee Bean' regularly invests in training and development programmes to equip employees with the necessary skill and knowledge. 'Coffee Bean' believed that efficient employees will produce more and earn more. This will ensure their greatest efficiency and prosperity for both company and workers. The above case highlights the use of Scientific principles and techniques by 'Coffee Bean'. Explain any one such principle and two techniques.
- CBSE CLASS XII - 2024
- Marketing Management
- Preeti started her own cooking channel on 'Youtube Mood Art'. As her subscribers increased, she was not in a position to manage everything on her own. She hired Rahul and Riya to help her with filming editing, lighting and content research. She granted authority to them to operate within prescribed limits. She was thus, able to use her time on high priority activities like developing new recipes and content development etc. As a result, Rahul and Riya were given opportunities to develop and exercise initiative. Preeti was now able to focus on objectives and meet the target of achieving a subscriber base of one million in six months.
(a) Identify the concept of management used by Preeti.
(b) Explain any five points of importance of the concept identified in .- CBSE CLASS XII - 2024
- Organising
- Explain the following types of plans:
(b) Strategy
(ii) Method
(iii) Budget- CBSE CLASS XII - 2024
- Principles of Management
View More Questions