# Calculate the number of molecules of phosphorus [P4] present in 31 gram of phosphorus.

Molecular mass of P4 = 124 g/mol.

i.e.

Mass of 1 mole ( 6.023 X 1023) P4 molecules = 124 g

Hence..

124 g  P4=6.023 x 1023  P4 molecules

so

1g P4 =( 6.023 x 1023)/(124) =0.25 x 1023molecules

and

31 g P4 =  0.25 x 1023 x 31 = 1.5 x1023 molecules.

• -4

Phosphorus molecule, P4
Atomic mass of phosphorus = 31 u
:. Molecular mass of P4 = 4 × 31 = 124 u
:. Molar mass of P4 is 124 u.

Given mass (m) of S8 molecule = 31 g
Avogadro's number (N0) = 6.023 × 1023

Number (N) of particles (molecules) = Given mass/ Molar mass ×Avogadro number

or N = m/ M ×N0

= N = 31/124 × 6.023 × 1023

=> N = 1.5057 × 1023

hence,

Number of molecules = 1.5057 *1023

• 1

Phosphorus molecule, P4
Atomic mass of phosphorus = 31 u
:. Molecular mass of P4 = 4 * 31 = 124 u
:. Molar mass of P4 is 124 u.

Given mass (m) of P4 molecule = 31 g
Avogadro 's number (N0) = 6.023 * 1023

Number (N) of particles (molecules) = Given mass/ Molar mass *Avogadro number

or N = m/ M *N

= N = 31/124* 6.023* 1023

=> N = 1.5057 * 1023

hence,

Number of molecules = 1.5057 *1023

• 23

Phosphorus molecule, P4
Atomic mass of phosphorus = 31 u
:. Molecular mass of P4 = 4 * 31 = 124 u
:. Molar mass of P4 is 124 u.

Given mass (m) of P4 molecule = 31 g
Avogadro 's number (N0) = 6.023 * 1023

Number (N) of particles (molecules) = Given mass/ Molar mass *Avogadro number

or N = m/ M *N

= N = 31/124* 6.023* 1023

=> N = 1.5057 * 1023

hence,

Number of molecules = 1.5057 *1023

• 2

Thank u all.

• 2
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