Explanation:
The given data is as follows.
Thickness = 0.28 mm =
cm = 0.028 cm
Area = 0.40
=
= 4000
As, it is known that volume = area × thickness
So, Volume =
= 112 ![cm^{3}](https://tex.z-dn.net/?f=cm%5E%7B3%7D)
As density is mass divided by volume. So, mass of chromium will be calculated as follows.
Density =
7.20
=
mass = 806.4 g
As, mass of 1 mole of chromium is 52 g. So, number of moles in 806.4 g of chromium will be as follows.
No. of moles =
= ![\frac{806.4 g}{52 g}](https://tex.z-dn.net/?f=%5Cfrac%7B806.4%20g%7D%7B52%20g%7D)
= 15.50 mol
In chromate ion, (
) charge on Cr is +6. It means that 6 electrons are needed to reduce
into Cr.
As, 1 mole of
ions require 6 moles of electrons. Therefore, moles of electrons for 15.50 mol will be calculated as follows.
6 × 15.50 mol = 93.04 mol
To calculate number of electrons we multiply number of moles by Avogadro's number as follows.
![93.04 mol \times 6.02 \times 10^{23}](https://tex.z-dn.net/?f=93.04%20mol%20%5Ctimes%206.02%20%5Ctimes%2010%5E%7B23%7D)
= ![560.13 \times 10^{23}](https://tex.z-dn.net/?f=560.13%20%5Ctimes%2010%5E%7B23%7D)
=
electrons
There is magnitude of
times the charge on an electron is equal to 1 coulomb.
Hence, number of coulombs will be as follows.
No. of coulombs = ![\frac{5.6 \times 10^{25}}{6.241 \times 10^{18}}](https://tex.z-dn.net/?f=%5Cfrac%7B5.6%20%5Ctimes%2010%5E%7B25%7D%7D%7B6.241%20%5Ctimes%2010%5E%7B18%7D%7D)
=
C
or, =
C
Thus, we can conclude that
C are required to plate a layer of chromium metal with given data.