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Two non reactive gases A and B are present in a container with partial pressure 200 and 180 mm of Hg. When a third non reactive gas C is added then total pressure becomes 1atm, then mole fraction of C will b a) 0.75 b) 0.5 c) 0.25 d) cannot b calculated?
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Two non reactive gases A and B are present in a container with partial...
Mole fraction of gas C can be calculated by using Dalton's law of partial pressures. According to this law, the total pressure exerted by a mixture of non-reactive gases is equal to the sum of the partial pressures of each gas.

1. Calculate the mole fraction of gases A and B:
- The partial pressure of gas A is given as 200 mm Hg.
- The partial pressure of gas B is given as 180 mm Hg.
- Total pressure is given as 1 atm.

2. Convert the partial pressures to atm:
- 1 atm = 760 mm Hg.
- Partial pressure of gas A = 200 mm Hg / 760 mm Hg/atm = 0.2632 atm.
- Partial pressure of gas B = 180 mm Hg / 760 mm Hg/atm = 0.2368 atm.

3. Calculate the mole fraction of gas A and B:
- Mole fraction of gas A = Partial pressure of gas A / Total pressure = 0.2632 atm / 1 atm = 0.2632.
- Mole fraction of gas B = Partial pressure of gas B / Total pressure = 0.2368 atm / 1 atm = 0.2368.

4. Subtract the mole fractions of gas A and B from 1 to find the mole fraction of gas C:
- Mole fraction of gas C = 1 - (Mole fraction of gas A + Mole fraction of gas B)
- Mole fraction of gas C = 1 - (0.2632 + 0.2368) = 1 - 0.5 = 0.5.

Therefore, the mole fraction of gas C is 0.5, which corresponds to option b) 0.5.

This calculation assumes that the gases A, B, and C are ideal gases and do not react with each other. It also assumes that the total pressure is equal to the sum of the partial pressures of each gas, as stated by Dalton's law.
Community Answer
Two non reactive gases A and B are present in a container with partial...
The answer is option B)0.5...!

P(total)= P(A)+P(B)+P(C)

P(A)=200mm=200/760--->0.26 Atmos

P(B)=180mm=180/760--->0.23

P(total)=1

P(C)=1-0.26+0.25--->0.51

therefore; approximately: P(C)=0.5...!hope u clear..!😊👍
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Two non reactive gases A and B are present in a container with partial pressure 200 and 180 mm of Hg. When a third non reactive gas C is added then total pressure becomes 1atm, then mole fraction of C will b a) 0.75 b) 0.5 c) 0.25 d) cannot b calculated?
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Two non reactive gases A and B are present in a container with partial pressure 200 and 180 mm of Hg. When a third non reactive gas C is added then total pressure becomes 1atm, then mole fraction of C will b a) 0.75 b) 0.5 c) 0.25 d) cannot b calculated? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Two non reactive gases A and B are present in a container with partial pressure 200 and 180 mm of Hg. When a third non reactive gas C is added then total pressure becomes 1atm, then mole fraction of C will b a) 0.75 b) 0.5 c) 0.25 d) cannot b calculated? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two non reactive gases A and B are present in a container with partial pressure 200 and 180 mm of Hg. When a third non reactive gas C is added then total pressure becomes 1atm, then mole fraction of C will b a) 0.75 b) 0.5 c) 0.25 d) cannot b calculated?.
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