Question

A silver ornament of mass '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.

Explain with complete calculations & justifications.

Points : 30 ☺

Answer 1

let atomic mass of Ag (silver)= Mg
and atomic mass of Au (gold) = Mu

now, mass of gold = 1% mass of silver
mass of gold = m/100 gram
now,
no of mole of gold = given wt/atomic mass
= m/100/Mu =m/100Mu
no of atoms of gold = mole of gold x Avogadro's constant ( A let)

=mA/100MuA -----------(1)

no of mole of silver = m/Mg

no of atoms of silver = mole of silver x A
=mA/Mg ---------(2)

no of atoms of gold /no of atoms of silver ={mA/100Mu}/{mA/Mg}
{ from eqn (1) and (2)

ratio = Mg/100Mu
=108/19700


put atomic mass of silver and atomic mass of gold.

Answer 2

Hi there!

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 × 10²³


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 × 10²³  :  m ÷ 108 × 6.022 × 10 ²³

6.022× 10²³ 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