A) \[1.5L\]
B) \[3.36L\]
C) \[1.05L\]
D) \[0.07L\]
Correct Answer: C
Solution :
Moles of \[{{N}_{2}}{{O}_{5}}=\frac{10.8}{108}=0.1\] and, \[n=\frac{9.6}{2.4}=4\] here n = numbers of half-life. \[{{N}_{2}}{{O}_{5}}\xrightarrow{{}}2N{{O}_{2}}+\frac{1}{2}{{O}_{2}}\] \[\therefore \] \[{{N}_{t}}=0.1\times {{\left( \frac{1}{2} \right)}^{n}}\] Moles of \[{{N}_{2}}{{O}_{5}}\] left \[=\frac{0.1}{16}\] Moles of \[{{N}_{2}}{{O}_{5}}\] changed to product \[=\left( 0.1-\frac{0.1}{16} \right)=\frac{1.5}{16}mol\] Moles of \[{{O}_{2}}\] formed \[=\frac{1.5}{16}\times \frac{1}{2}=\frac{1.5}{32}\] Volume of oxygen \[=\frac{1.5}{32}\times 22.4\] \[=1.05L\]
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