A) 0.6 atm
B) 1.2 atm
C) 2.4 atm
D) 3.6 atm
Correct Answer: B
Solution :
Partial pressure exerted by any gas in a mixture of non reacting gases (p) \[p=\frac{moles\,of\,that\,gas}{total\,moles}\times total\,pressure\] let x g of all gases are mixed mol of \[S{{O}_{2}}=\frac{x}{64}\] mol of \[C{{H}_{4}}=\frac{x}{16}\] mol of \[{{O}_{2}}=\frac{x}{32}\] [\[\because \] molecular mass of \[S{{O}_{2}},C{{H}_{4}},{{O}_{2}}\] are 64, 16, 32 respectively] Total moles \[=\frac{x}{64}+\frac{x}{16}+\frac{x}{32}\] \[=\frac{x+4x+2x}{64}=\frac{7x}{64}\] \[{{p}_{C{{H}_{4}}}}=\frac{x}{\frac{16}{\frac{7x}{64}}}\times 2.1=1.2atm\]
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