Question

A silver ornament of m gram 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.

• A. 108 : 19700
• B. 54 : 19700
• C. 108 : 197
• D. 10.8 : 197

Answer 1

The correct option is A 108 : 19700

Mass of silver$=$mg
Mass of gold $=\frac{m}{100}g$
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
$=\frac{m}{100\times 197}\times N_A:\frac{m}{108}\times N_A$
$108:100\times 197$
$=108:19700$
$=1:182.41$