Equal masses of CH4 and O2 are mixed in an empty container at 25°C. The pressure exerted by oxygen is
A)16/17
B) 1/9
C) 1/3
D) 8/9
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The pressure exerted by oxygen is 1/9
Amulayaz1:
How?
@nishtha111 can you tell me the complete method?
For CH4: 16.04 g/mol≈16 g/mol For H2: 2.016 g/mol≈2 g/molFor CH4:mg⋅1 mole CH416g=m16 moles CH4
For H2:mg⋅1 mole H22g=m2 moles H2
The total number of moles will be
ntotal=m16+m2=m⋅(16+2)16⋅2
The mole fraction of methane will once again be
χCH4=m16⋅16⋅2m⋅(16+2)=218=
For H2:mg⋅1 mole H22g=m2 moles H2
The total number of moles will be
ntotal=m16+m2=m⋅(16+2)16⋅2
The mole fraction of methane will once again be
χCH4=m16⋅16⋅2m⋅(16+2)=218=
For CH4:mg⋅1 mole CH4/16g=m/16 moles CH4
For H2:mg⋅1 mole H2/2g=m/2 moles H2
The total number of moles will be
ntotal=m/16+m/2=m⋅(16+2)/16⋅2
The mole fraction of methane will once again be
χCH4=m/16⋅16⋅2/m⋅(16+2)=2/18=
1/9
For H2:mg⋅1 mole H2/2g=m/2 moles H2
The total number of moles will be
ntotal=m/16+m/2=m⋅(16+2)/16⋅2
The mole fraction of methane will once again be
χCH4=m/16⋅16⋅2/m⋅(16+2)=2/18=
1/9
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