Platinum nanoparticles are being explored for use as high-performance co-catalys

Question

Platinum nanoparticles are being explored for use as high-performance co-catalysts in sunlight driven photocatalysis of water and various organic species. These particles greatly enhanced the catalytic activity of the primary photon absorbing semiconductors they decorate. Successful work in this field could lead to the ability to produce clean burning fuels from water or waste CO_2. As part of a research program you are concerned about the total number of free electrons found in a specific Pt nanoparticle. Assuming that density of Pt is 21.45 g/cm^3 and that Pt forms cubic nanoparticles with length 3.92 nm, calculate how many atoms would be in such a particle? Also how many free electrons are present in an ideal particle assuming Pt commonly has a 2+ valence state?

Explanation / Answer

d= densityof Pt = 21.45 g cm3

a= length of cube side of Pt nanoparticle = 3.92 nm

thus V = volume of cube = a3 = (3.92)3 = 60.24 nm3

since, 1 nm = 10-7 cm

V = 60.24 * 10-21 cm3

density = mass volume

d = mV

21.45 g cm3 = m0.24 * 10-21 cm3

m = 1292.15*10-21 g = 1.3*10-24 g

Theoritical mass of Pt = 195 g

i.e 195 g of Pt contains 6.023*10-23 atoms

thus, 1.3*10-24 g of Pt contains = 6.023*10-23 atoms * (1.3*10-24 g ) 195 g = 4.02* 10-49 atoms

thus, Pt nanoparticles contains 4.02* 10-49 atoms

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