A) \[2.88\times {{10}^{-3}}\]
B) \[28.8\times {{10}^{-3}}\]
C) \[288\times {{10}^{-3}}\]
D) \[28.8\times {{10}^{3}}\]
Correct Answer: A
Solution :
\[6.023\times {{10}^{23}}\]molecules = 1 mole of \[C{{O}_{2}}=44g\] \[{{10}^{21}}\] molecules of \[C{{O}_{2}}=\frac{44\times {{10}^{21}}}{6.023\times {{10}^{23}}}g\] \[=7.31\times {{10}^{-2}}g\] \[=7.31mg\] \[\therefore \] \[C{{O}_{2}}\] left \[=200-73.1=126.9\text{ }mg\] Hence, moles of \[C{{O}_{2}}\], left \[=\frac{given\,mass}{mol.\,mass}\] \[=\frac{126.9\times {{10}^{-3}}}{44}\] \[=2.88\times {{10}^{-3}}mol\]You need to login to perform this action.
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