A) 24 g of C
B) 56 g of Fe
C) 26 g of Al
D) 108 g of Ag
Correct Answer: A
Solution :
No. of atoms in 24 g of\[C=\frac{24}{12}\times 6.02\times {{10}^{23}}\] \[=2\times 6.02\times {{10}^{23}}\] No. of atoms in 56 g of \[Fe=\frac{56}{56}\times 6.02\times {{10}^{23}}\] No. of atoms in 26 g of Al \[=\frac{26}{27}\times 6.02\times {{10}^{23}}\] \[\approx 6.02\times {{10}^{23}}\] No. of atoms in 108 g of \[Ag=\frac{108}{108}\times 6.02\times {{10}^{23}}\] \[=6.02\times {{10}^{23}}\]
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