A) \[\operatorname{L}\,\,of\,\,{{H}_{2}}\,gas\,STP\]
B) \[5L\,\,of {{N}_{3}} gas STP\]
C) \[0.5g of {{H}_{2}} gas\]
D) \[10g of {{O}_{2}} gas\]
Correct Answer: A
Solution :
\[15L of {{H}_{2}} gas STP\] \[\operatorname{At}\,STP,\] \[22.4 L of any gas = 6.02\times 1023 molecules\] \[\therefore 15\operatorname{L}\,\,{{\operatorname{H}}_{2}}=\frac{6.02\times {{10}^{23}}}{22.4}\times 15=4.03\times {{10}^{23}}\] \[\therefore 5\operatorname{L}\,\,{{\operatorname{N}}_{2}}=\frac{6.02\times {{10}^{23}}}{22.4}\times 15=1.344\times {{10}^{23}}\] \[2\operatorname{g}\,\,{{H}_{2}}=6.02\times 1{{0}^{23}}molecules\] \[\therefore 0.5\operatorname{g}\,\,{{H}_{2}}=\frac{6.02\times {{10}^{23}}}{2}\times 0.5\] \[=1.505\times 1{{0}^{23}}\,\,\,molecules\] \[32 g {{O}_{2}}= 6.02\times 1023 molecules\] \[\therefore 10 g {{O}_{2}}= \frac{6.02\times 1023}{32}\times 10\] \[=1.88\times 10=23 molecules\]
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