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A coating of coabalt that is 0.005 cm th...

A coating of coabalt that is 0.005 cm thick is deposited on a plate that is `0.5 m^(2)` in total area. How many atoms of cobalt are deposited on the plate ? (Density of cobalt `= 8.9 cm^(-3)` , Atomic mass of Co = 59 amu)

Text Solution

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Area of the plate `= 0.5 m^(2) = 0.5 xx 10^(4) cm^(2)`
Thickness of the coating `= 0.005 cm`
Volume of cobalt deposited `= 0.5 xx 10^(4) xx 0.005 = 25 cm^(3)`
Mass of cobalt deposited = Volume `xx` density
`= 25 cm^(3) xx 8.9 g cm^(-3) = 222.5 g`
Gram atomic mass of cobalt = 59 g
Now, 59 g of cobalt contain atoms `= 6.022 xx 10^(23)`
`222.5 g` of cobalt will contain atoms `= (6.022xx10^(23))/((59g))xx(222.5g)=2.27xx10^(24)` atoms.
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