A) 100
B) \[6.02\times {{10}^{23}}\]
C) \[6.02\times {{10}^{24}}\]
D) \[6.02\times {{10}^{25}}\]
Correct Answer: D
Solution :
Amount of gold\[=19.7kg\]\[=19.7\times 1000gm\]=19700gm No. of moles \[=\frac{19700}{197}=100\] \[\therefore \] No. of atoms \[=100\times 6.023\times {{10}^{23}}\] \[=6.023\times {{10}^{25}}\]atoms.You need to login to perform this action.
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