A) \[6\times {{10}^{20}}\]
B) \[1.2\times {{10}^{21}}\]
C) \[3\times {{10}^{21}}\]
D) \[3.6\times {{10}^{21}}\]
Correct Answer: D
Solution :
Mol. wt of \[{{C}_{2}}{{H}_{5}}OH\] \[=2\times 12+5+16+1=64\] \[\because \] \[48g\,{{C}_{2}}{{H}_{5}}OH\] has H atom \[=6\times {{N}_{A}}\] \[\therefore \] \[0.046g\,\,{{C}_{2}}{{H}_{5}}OH\] has H atoms \[=\frac{6\times 6.02\times {{10}^{23}}\times 0.046}{46}\]\[=3.6\times {{10}^{21}}\]
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