A) \[1.33\]
B) \[2.66\]
C) \[2.00\]
D) \[3.00\]
Correct Answer: A
Solution :
[a] Excess \[AgN{{O}_{3}}\] gave \[0.25\text{ }mol\]of yellow precipitate. \[\Rightarrow \,\,\,\,\,\,0.5-x=0.25\] \[\Rightarrow \,\,\,\,\,\,x=0.25\] \[=\frac{0.25/1}{\left( \frac{0.75}{1} \right)\left( \frac{0.25}{1} \right)}=1.33\,(V=1.0L)\]You need to login to perform this action.
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