Question

The number of silver atoms present in a 90% pure silver wire weighing $10g$ is:

• A. $5.57\times {10}^{22}$
• B. $0.62\times {10}^{23}$
• C. $5.0\times {10}^{22}$
• D. $6.2\times {10}^{29}$

Answer 1

The correct option is C $5.0\times {10}^{22}$

Mass of pure silver= $\frac{90}{100}\times 10g=9g$

$\therefore$ Moles of silver=$\frac{9}{108}=0.083$

$\therefore$ Number of silver atoms=$0.083\times 6.022\times {10}^{23}$

=$0.5\times {10}^{23}$

=$5\times {10}^{22}$