A) \[6.023\times {{10}^{23}}\]
B) 2
C) 1
D) \[12.046\times {{10}^{23}}\]
Correct Answer: D
Solution :
\[\because \]\[23\,g\]of Na contains \[=6.023\times {{10}^{23}}\,\]atoms \[\therefore \]1 g of Na contains \[=\frac{6.023\times {{10}^{23}}}{23}\]atoms \[\therefore \]46 g of Na contains \[=\frac{6.023\times {{10}^{23}}}{23}\times 46\]atoms \[=12.046\times {{10}^{23}}\,\]atoms
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