A) \[7\,g\,{{N}_{2}}\]
B) \[2\,g\,{{H}_{2}}\]
C) \[16\,g\,N{{O}_{2}}\]
D) \[16\,g\,{{O}_{2}}\]
Correct Answer: B
Solution :
In 7g nitrogen, number of molecules \[=\frac{7.0}{28}\,mol=0.25\times N\,\,molecules\] where N = Avogadro number \[=6.023\times {{10}^{23}}\] \[In\,2\,g\,\,{{H}_{2}}=\frac{2.0}{2}mol=1\times N\,molecules\] \[In\,16\,g\,N{{O}_{2}}=\frac{16.0}{46}mol=0.348\,\times \,N\,\,molecules\] 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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