A) \[6.02\times {{10}^{23}}\times 4\]
B) \[6.02\times {{10}^{23}}\]
C) \[\frac{6.02\times {{10}^{23}}}{2}\]
D) \[\frac{6.02\times {{10}^{23}}}{4}\]
Correct Answer: D
Solution :
Sol. Number of atoms in 1 g of helium \[=\frac{Avogadros\,number}{weight}\] \[=\frac{6.02\times {{10}^{23}}}{4}\]
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