A) 0.01 g of hydrogen
B) 0.085 g of \[N{{H}_{3}}\]
C) 320 mg of gaseous \[S{{O}_{2}}\]
D) All of the above
Correct Answer: D
Solution :
\[Moles=\frac{weight}{molecular\,weight}\] \[=\frac{volume\,(in\,L.)}{22.4}\] \[\therefore \]\[\frac{weight}{molecular\,weight}=\frac{molecular\,weight}{weight\times 22.4}\] or \[\frac{1}{Volume\,(in\,L)}=\frac{molecular\,weight}{weight\times 22.4}\] For \[0.22g\,C{{O}_{2}},\] \[\frac{1}{Volume}=\frac{44}{0.22\times 22.4}=\frac{200}{22.4}\] For 0.01 g of hydrogen, \[\frac{1}{Volume}=\frac{2}{0.01\times 22.4}=\frac{200}{22.4}\] For \[0.085g\,\,N{{H}_{3}}\], \[\frac{1}{Volume}=\frac{17}{0.085\times 22.4}\] \[=\frac{200}{22.4}\] For 0.32 g gaseous \[S{{O}_{2}}\], \[\frac{1}{Volume}=\frac{64}{0.32\times 22.4}\] \[=\frac{200}{22.4}\] Thus, all have the same volume at NTP.
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