A) 142mm
B) 156mm
C) 182mm
D) 180mm
Correct Answer: D
Solution :
\[56g{{N}_{2}}=\frac{58}{28}=2\,mol\] \[44gC{{O}_{2}}=\frac{44}{44}=1\,mol\] \[16gC{{H}_{4}}=\frac{16}{16}=1\,mol\] Partial pressure of \[C{{H}_{4}}\] \[=\frac{{{n}_{C{{H}_{4}}}}}{{{n}_{{{N}_{2}}}}+{{n}_{C{{O}_{2}}}}+{{n}_{C{{H}_{4}}}}}\times p\] \[=\frac{1}{2+1+1}\times 720=180\,mm\]You need to login to perform this action.
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