A silver ornament of mass m gram is polished with gold equivalent to 1...
Assume that mass of silver to be "m" grams.
It is given that mass of gold = 1% of mass of silver
That is,
mass of gold = 1 % of m grams
Mass of gold = 0.01 m grams [ 1% = 1 / 100 = 0.01]
Molar mass of Gold (Au) = 197 g
Given mass = 0.01 m grams
No : of moles = Given mass / Molar mass
= 0.01 m / 197 moles
No : of atoms = No : of moles * Avogadro no:
= [ 0.01 m / 197 ] * 6.022 * 10^23
Molar mass of silver (Ag) = 108 g
Given mass = m grams
No: of moles = Given mass / Molar mass
= m / 108 moles
No : of atoms = No : of moles * Avogadro no:
= m / 108 / 6.022 / 10 ^23
Ratio of no : of atoms of gold and silver :-
0.01 m / 197 * 6.022 * 10^23 : m / 108 * 6.022 * 10 ^23
6.022 * 10^23 will get cancelled on both sides.
0.01 m / 197 = m / 108
Cross multiplication is done.
0.01 * 108 : 197 * m
1.08 = 197 m
m = 1.08 / 197
= 0.0054 (approx.)
Therefore ratio of number of atoms of gold and silver in the ornament :
0.0054 : 1
OR
5.4 * 10^-5 : 1