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kylabreanne120

30.04.2022

Chemistry

Time required to deposit an even layer of gold with given thickness is $5.3 \times 10^{2}$ sec.

Explanation:

The given data is as follows.

Surface area = 49.8 $cm^{2}$ ,

Density of gold = 19.3 $g/cm^{3}$ ,

Current = 3.15 A, thickness of gold layer = $1.2 \times 10^{-3} cm$

It is known that relation between volume, area and thickness is as follows.

V = Surface area × Thickness

= $49.8 \times 1.2 \times 10^{-3} cm$

= 0.05988 $cm^{3}$

Therefore, we will calculate the time required to deposit an even layer of gold with given thickness is calculated as follows.

$0.05988 \times cm^{3} \times \frac{19.3 g Au}{1 cm^{3}} \times \frac{1 mol Au}{197 g Au} \times \frac{3 mol e^{-}}{1 mol Au} \times \frac{96485}{1 mol e^{-}} \times \frac{1 As}{1 C} \times \frac{1}{3.20 A}$

= $5.3 \times 10^{2}$ sec

Thus, we can conclude that time required to deposit an even layer of gold with given thickness is $5.3 \times 10^{2}$ sec.

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