Question

Question 42
A silver ornament of mass ‘m’ gram is polished with gold equivalent to $1\%$ of the mass of silver. Compute the ration of the number of atoms of gold and silver in the ornament.

Answer 1

Mass of silver = m g
Mass of gold $=\frac{m}{100}g$
Atomic mass of Ag = 108 u
Atomic mass of Au = 197 u
Number of atoms of silver $=\frac{Mass}{Atomicmass}\times N_A$$=\frac{m}{108}\times N_A$
Number of atoms of gold $=\frac{m}{100\times 197}\times N_A$
Ratio of number of atoms of gold to silver
$=Au:Ag=\frac{m}{100\times 197}\times N_A:\frac{m}{108}\times N_A$
$=108:100\times 197=108:19700=1:182.41$