A) 22.4
B) 2.24
C) 20.16
D) 2.016
Correct Answer: D
Solution :
Amount of pure lime stone \[(CaC{{O}_{3}})\] is 10g of 90% sample \[=\frac{90}{100}\times 10=9\,g\] Molecular mass of \[CaC{{O}_{3}}=100\] Thus, 100 g of lime stone gives 22.4 L of \[C{{O}_{2}}\] at STP 9 g of lime stone gives 22.4 L of \[C{{O}_{2}}\] at STP. 9 g of lime stone will give\[=\frac{22.4}{100}\times 9=2.016\,L\]You need to login to perform this action.
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