Question

The number of atoms present in $10.8\, g$ of silver is :
[atomic weight of silver $=108$ ]

• A. $ 6.123\times {{10}^{23}} $
• B. $ 6.312\times {{10}^{22}} $
• C. $ 6.023\times {{10}^{22}} $
• D. $ 6.076\times {{10}^{23}} $

Answer 1

Answer: (C)

$\because 108\,g$ of silver contains $=6.02 \times 10^{23}$ atoms of $Ag$
$\therefore 10.8 \,g $ of silver contains $=\frac{6.023 \times 10^{23}}{108} \times 10.8 $
$=6.023 \times 10^{22} $ atoms