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

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