A) 45.215 atm
B) 31.205 atm
C) 25.215 atm
D) 15.210 atm.
Correct Answer: C
Solution :
Moles of oxygen, \[^{n}{{o}_{2}}=\frac{4}{32}=0.125mol\] Moles of Hydrogen, \[^{n}{{H}_{2}}=\frac{2}{2}=1.00mol\] Temperature = 273 K, volume = 1 Litre \[\Rightarrow \] Total pressure \[=\frac{({{n}_{O}}_{_{2}}+{{n}_{{{H}_{2}}}})\times RT}{V}\] \[=\frac{1.125\times 0.0821\times 273}{1}\] \[=25.215atm.\]You need to login to perform this action.
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