A) 25.215 atm
B) 31.205 atm
C) 45.215 atm
D) 15.210 atm
Correct Answer: A
Solution :
no. of moles of \[{{O}_{2}}=\frac{4}{32}=0.125\] no. of moles of \[{{H}_{2}}=\frac{2}{2}=1\] total no. of moles = \[1+0.125=1.125\] \[P=\frac{nRT}{V}=\frac{1.125\times 0.082\times 273}{1}=25.184atm.\]You need to login to perform this action.
You will be redirected in
3 sec