Question

What is the difference in mass between 3.01×10^24 atoms of gold and a gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm

Answer 1

Answer:

The difference in mass between 3.01×10^24 atoms of gold and a gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm is :

Difference  in mass = 985.32 - 984.5 = 0.82 g

Explanation:

Part I :

$n =\frac{3.01\times 10^{24}}{6.022\times 10^{23}}$

n = 4.9983

n = 4.99 moles

(Note : You can also take n = 5 mole )

Molar mass of gold = 196.96 g/mole

This means, 1 mole of gold(Au) contain = 196.96 grams

So, 4.99 moles of gold contain = $5\times 196.96$ g

4.99 moles of gold contain = 984.8 g

Mass of ${3.01\times 10^{24}}$ atoms of gold = 984.5 g

Part II :

Density of Gold = $19.32 g/cm^{3}$

Volume of the cuboid = $length\times breadth\times height$

Volume of the gold bar =$6.00\times 4.25\times 2.00$

Volume of the gold bar = 51$cm^{3}$

Using formula,

$Density = \frac{mass}{Volume}$

$Mass = Density\times Volume$

$Mass = 19.32 \times 51$

Mass = 985.32 g

So, A  gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm has mass of 985.32 g

Difference  in mass = 985.32 - 984.5 = 0.82 g