A) 26%
B) 40%
C) 58.33%
D) 66.66%
Correct Answer: C
Solution :
[c] \[P{{o}_{2}}=\frac{3}{10}\times {{P}_{T}};\] After removing 2 mole of \[{{O}_{2}}\] \[P{{'}_{{{O}_{2}}}}=\frac{1}{10}\times {{P}_{T}}\] Decrease in partial pressure of \[{{O}_{2}}\] \[\frac{\frac{3{{P}_{T}}}{10}-\frac{{{P}_{T}}}{8}}{\frac{3{{P}_{T}}}{10}}\times 100=58.33%\]%You need to login to perform this action.
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