Chemical Kinetics and Nuclear ChemistryHard
Question
the density of gold is 19g/cm3. If 1.9 × 10-4g of gold is dispersed in one litre of water to give a sol having spherical gold particles of radius 10 nm, then the number of gold par tiles per mm3 of the so will be
Options
A.1.9 × 1012
B.6.3 × 1014
C.6.3 × 1010
D.2.4 × 106
Solution
Volume of gold present in solution
= 0.1 × 10-4 cm3
For spherical particle of gold with radius equal to 10mn The volume of each particle
=
πr3 = 
×(10 × 10-7cm)3
=
×
Number of gold particle present
× 1013 particles
= 2.4 × 1012 particles
= 2.4 × 1012 particles of gold are present in 1000cm3 (1litre).
∴ Number of particles present per mm3
[1L = 106 mm3]
= 2.4 × 106
= 0.1 × 10-4 cm3
For spherical particle of gold with radius equal to 10mn The volume of each particle
=
πr3 = ×(10 × 10-7cm)3=
× Number of gold particle present
× 1013 particles = 2.4 × 1012 particles
= 2.4 × 1012 particles of gold are present in 1000cm3 (1litre).
∴ Number of particles present per mm3
[1L = 106 mm3]= 2.4 × 106
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