A) \[75\,\,atm\]
B) \[160\,\,atm\]
C) \[180\,\,atm\]
D) \[215\,\,atm\]
Correct Answer: C
Solution :
First of all we have to calculate the number of moles. Number of moles of\[{{N}_{2}}=\frac{56}{28}=2\] Number of moles of\[C{{O}_{2}}=\frac{44}{44}=1\] Number of moles of\[C{{H}_{4}}=\frac{16}{16}=1\] \[\therefore \]Total number of moles\[=2+1+1=4\] \[\therefore \]mole fraction of\[C{{H}_{4}}\] \[\therefore \]partial pressure of \[C{{H}_{4}}\] = mole fraction of \[C{{H}_{4}}\times \] total pressure \[=\frac{1}{4}\times 720=180\,\,atm\]
You need to login to perform this action.
You will be redirected in
3 sec
