Question

A silver ornament of mass "m" grams is polished with gold equivalent to 1% of the mass of silver. Then the ratio of number of atoms of gold and silver in the ornament is

• A. 108 : 197
• B. 1 : 182.41
• C. 108 : 19700
• D. Both 2 and 3

Answer 1

Answer: (C)

108 : 19700

Mass of silver$=mg$---------------1

Mass of gold$=\frac{m}{100}g$--------------2

Number of atoms of $Ag=\frac{m}{108\times N_A}$ (Atomic Weight of $Ag=108$)

Number of atoms of $Au=\frac{m}{100\times 197\times N_A}$

$Au:Ag\equiv \frac{m}{100\times 197\times N_A}:\frac{m}{108\times N_A}$

$=108:100\times 197$

$=108:19700$