A) \[220\,mm\]
B) \[540\text{ }mm\]
C) \[180\text{ }mm\]
D) \[100\text{ }mm\]
Correct Answer: C
Solution :
\[56\,g\,\,{{N}_{2}}=\frac{56}{28}=2\,mol\] \[44\,g\,\,C{{O}_{2}}=\frac{44}{44}=1\,mol\] \[16\,g\,\,C{{H}_{4}}=\frac{16}{16}=1\,mol\] 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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