A) \[2.24\text{ }L\]
B) \[1.12L\]
C) \[0.84\text{ }L\]
D) \[0.56\text{ }L\]
Correct Answer: B
Solution :
\[BaC{{O}_{3}}\xrightarrow{\Delta }BaO+C{{O}_{2}}\] Molecular weight of \[BaC{{O}_{3}}=137+12+16\times 3\] \[=197u.\] \[\because \] \[197\text{ }g\text{ }BaC{{O}_{3}}\]gives 2\[22.4L\] \[C{{O}_{2}}\] at NTP \[\therefore \] \[~9.85\text{ }g\text{ }BaC{{O}_{3}}\]gives \[\frac{22.4}{197}\times 9.85L\] \[C{{O}_{2}}\]at NTP Volume of \[C{{O}_{2}}=1.12L\]
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