A) \[7\,g\,{{N}_{2}}\]
B) \[7\,g\,{{H}_{2}}\]
C) \[16\,g\,N{{O}_{2}}\]
D) \[16\,g\,{{O}_{2}}\]
Correct Answer: B
Solution :
| In 7g nitrogen, number of molecules |
| \[=\frac{7.0}{28}\,mol=0.25\times N\,\,\text{molecules}\] |
| where \[N=\] Avogadro number \[=6.023\times {{10}^{23}}~\] |
| In \[2\,g\,\,{{H}_{2}}=\frac{2.0}{2}mol=1\times N\,\text{molecules}\] |
| In \[16\,g\,N{{O}_{2}}=\frac{16.0}{46}mol=0.348\,\times \,N\,\,\text{molecules}\] |
| In \[16\,g\,{{O}_{2}}=\frac{16}{32}\,mol=0.5\times N\,\,\text{molecules}\] |
| Hence, maximum molecules are present in 2g \[{{H}_{2}}\]. |
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