A) 22 g
B) 32 g
C) 44 g
D) 88 g
Correct Answer: A
Solution :
Key Idea Volume (STP) = mole \[\times 22.4\,L\] The volume of one mole of gas at STP = 22.4 L Mass of 1 mole of \[C{{O}_{2}}=12+2\times 16\] = 44 g \[\because \] The weight of 22.4 L of \[C{{O}_{2}}\] at STP = 44 g So, the weight of 11.2 L of \[C{{O}_{2}}\] at STP \[=\frac{44}{22.4}\times 11.2=22g\] Note- 11.2 L = mole \[\times 22.4\,L\] Mole \[=\frac{11.2}{22.4}=\frac{1}{2}\] \[\because \] The mass of 1 mole \[C{{O}_{2}}=44\,g\] \[\therefore \] The mass of \[\frac{1}{2}\] mole\[C{{O}_{2}}=\frac{44}{2}=22\,g\]
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