A) \[3.0\times {{10}^{23}}\]
B) \[6.02\times {{10}^{23}}\]
C) \[\frac{16}{6.02}\times {{10}^{23}}\]
D) \[\frac{16}{3.0}\times {{10}^{23}}\]
Correct Answer: B
Solution :
16gm of \[C{{H}_{4}}\]= 1mole \[=6.023\times {{10}^{23}}\] molecules.You need to login to perform this action.
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