A) 0.84 L
B) 2.24 L
C) 4.06 L
D) 1.12 L
Correct Answer: D
Solution :
\[BaC{{O}_{3}}\to BaO+C{{O}_{2}}\uparrow \] Molecular weight of \[BaC{{O}_{3}}=137+12+3\times 16\]=197 \[\because \] 197gm produces 22.4L at S.T.P. \[\therefore \] 9.85gm produces \[\frac{22.4}{197}\times 9.85=1.12L\]at S.T.P.You need to login to perform this action.
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