A silver ornament of mass m , g is polished w | vedclass

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(B) Mass of silver $= m \, g$Mass of gold $= \frac{m}{100} \, g$Number of atoms of silver $= \frac{	ext{mass}}{	ext{atomic mass}} 	imes N_A = \frac

Vedclass · Accepted Answer

$1: 182.41$

Vedclass · Answer

(B) Mass of silver $= m \, g$Mass of gold $= \frac{m}{100} \, g$Number of atoms of silver $= \frac{	ext{mass}}{	ext{atomic mass}} 	imes N_A = \frac{m}{108} 	imes N_A$Number of atoms of gold $= \frac{	ext{mass}}{	ext{atomic mass}} 	imes N_A = \frac{m}{100 	imes 197} 	imes N_A$Ratio of number of atoms of gold to silver $= \frac{	ext{Number of atoms of gold}}{	ext{Number of atoms of silver}}$$= \frac{\frac{m}{100 	imes 197} 	imes N_A}{\frac{m}{108} 	imes N_A} = \frac{108}{100 	imes 197}$$= \frac{108}{19700} \approx \frac{1}{182.41}$Thus,the ratio is $1: 182.41$.

$A$ silver ornament of mass $m \, g$ is polished with gold equivalent to $1\%$ of the mass of silver. Compute the ratio of the number of atoms of gold and silver in the ornament.

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