Question

$6.022\times {10}^{20}$ atoms of silver (at mass $=108u)$ weigh

• A. $108\times {10}^{3}g$
• B. $108g$
• C. $0.108g$
• D. $10.8g$

Answer 1

The correct option is C $0.108g$

Gram atomic mass of silver $=108g$
Since $108g$ of $Ag$ is equal to $1mole$ of $Ag$
Therefore $6.023\times {10}^{23}$ atoms of $Ag$ are contained by $=108g$ of $Ag$
$6.023\times {10}^{23}$ atoms will be contained by $\frac{108}{6.023\times {10}^{23}}\times 6.023\times {10}^{23}=0.108g$