A) \[1.4\times {{10}^{-21}}g\]
B) \[5.025\times {{10}^{-23}}g\]
C) \[1.3\times {{10}^{-19}}g\]
D) \[2.30\times {{10}^{-22}}g\]
Correct Answer: A
Solution :
Molecular weight of \[{{C}_{60}}{{H}_{122}}=(12\times 60)+(1\times 122)\] \[=720+122=842\] \[\therefore \] weight of 1 molecule of \[{{C}_{60}}{{H}_{122}}=\frac{\text{molecular}\,\text{weight}}{\text{Avogadro }\!\!\!\!\text{ s}\,\text{number}}\] \[=\frac{842}{6.0223\times {{10}^{23}}}\] \[=1.4\times {{10}^{-21}}g\]
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