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Equal weights of methane and oxygen are mixed in an empty container at 25°C. The fraction of the total pressure exerted by oxygen is a. 1/3 b. 1/2 c. 2/3 d. 1/3 × 273/298

Equal weights of methane and oxygen are mixed in an empty container at 25°C. The fraction of the total pressure exerted by oxygen is
a. 1/3
b. 1/2
c. 2/3
d. 1/3 × 273/298

Grade:upto college level

3 Answers

Aditi Chauhan
askIITians Faculty 396 Points
9 years ago
Sol. If x g of both oxygen and methane are mixed then : Mole of oxygen = x/32 Mole of methane = x/16 ⇒ mole fraction of oxygen = (x/32)/(x/32 + x/16) = 1/3 According to law of partial pressure Partial pressure of oxygen (po2) = mole-fraction × total pressure ⇒ 〖Po〗_2/p = 1/3
Shivani singh
15 Points
5 years ago
  1. Suppose the mass of methane as well as oxygen =w=1
  2. (Xo2) = w/32/w/32+w/16 = 1/32/3/32 = 1/3 
  3. Let the total pressure = p 
  4. Pressure exerted by oxygen =  Xo2 × Ptotal =
  5. P× 1/3
 
Yash Chourasiya
askIITians Faculty 256 Points
3 years ago
Dear Student

Equal masses of methane and oxygen are mixed in an empty container at25oC. The fraction of the total pressure exerted by oxygen is 1/3.

Let 32 g of oxygen are mixed with 32 g of methane.
The molar masses of methane and oxygen are 16 g/mol and 32 g/mole respectively
The number of moles of methane= 32g/16g ​= 2moles.
The number of moles of oxygen= 32g/32g​ = 1mole.
Total number of moles= 2+1 = 3.
The mole fraction of oxygen= 1/(1+2) ​= 1/3​.
The mole fraction is proportional to the fraction of the total pressure exerted.
Hence, the fraction of the total pressure exerted by oxygen is 1/3.

I hope this answer will help you.
Thanks & Regards
Yash Chourasiya

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