A) \[7g{{N}_{2}}\]
B) \[2g{{H}_{2}}\]
C) \[16gN{{O}_{2}}\]
D) \[16g{{O}_{2}}\]
Correct Answer: B
Solution :
In \[7g\] nitrogen, number of molecules \[=\frac{7.0}{28}mol=0.25\times N\]molecules where \[N=\] Avogadro number\[=6.023\times {{10}^{23}}\] In \[2\,\,g\,\,{{H}_{2}}=\frac{2.0}{2}mol=1\times N\]molecules In \[16\,\,g\,\,{{O}_{2}}=\frac{16}{32}mol=0.5\times N\]molecules In \[16\,\,g\,\,{{O}_{2}}=\frac{16}{32}mol=0.5\times N\]molecules Hence, maximum molecules are present in\[2\,\,g,\,\,{{H}_{2}}\].
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