Question

A silver ornament of mass 'm' gram is polished with gold equivalent to 1 percent of the mass of silver. Compare the ratio of the number of atoms of gold and silver in the ornament.

Answer 1

Let mass of the silver = m g

Mass of gold = 1% of the mass of silver = 1% of m g = 0.01m g.

Molar mass of silver =108 g/mol

Number of atoms in m g of silver = m108×NA
Molar mass of gold=197 g/mol

Number of atoms in 0.01m g of gold= 0.01m197×NA

Ratio of the number of atoms of gold to silver = 0.01m197×NA ×108m × NA = 5.4 × 10−31

Answer 2

FOR SILVER
Mass = m
no. of moles= m/108
no. of atoms= m/108×A
(Here A is the Avogadro's constant)
FOR GOLD
Mass = m/100
no. of moles= m/100×197
=m/19700
no. of atoms= m/19700×A

Ratio of gold atoms to silver atoms=
(m/19700×A)/(m/108×A)
= 108/19700
or(0.0054:1)