Question

Rama’s father who is a science teacher tried to teach her mole concept. For this she told
her that as there are 12 bananas in one dozen, there are 6.02x10 23 particles in a mole
which is equivalent to molar mass of an element. He told suppose a 0.005 cm tick coating
of gold is deposited on a medal of 0.5m2 total area the density of gold is 19.30g cm-3and
its atomic mass is 196.97 , then we can calculate exactly how many atoms of gold are
deposited.
(i) What is the mass of gold atoms deposited on the medal?
(ii) Calculate the number of moles of gold deposited on the medal.
(iii) Calculate the number of gold atoms deposited on the medal
(iv) What will be the volume occupied by one mole of a gas at STP?



PLS FIND THE ANSWER ASAP......

Answer 1

Given :

Thickness of gold layer = 0.005 cm

Area of medal = 0.5 m²

• 0.5 m² = 0.5 × 10⁴ cm²

Density of gold = 19.30 g/cm³

To Find :

i) Mass of gold atoms deposited on the medal.

ii) Number of moles of gold deposited on the medal.

iii) Number of gold atoms deposited on the medal.

iv) Volume occupied by one mole of a gas at STP.

Solution :

Are you scared after seeing length of the question? Question is just based on basic knowledge of mole concept.

Let's solve it step by step! :)

A] Mass of gold atoms :

❒ Density is defined as the ratio of mass to the volume.

$\sf:\implies\:Density=\dfrac{Mass}{Volume}$

$\sf:\implies\:Density=\dfrac{Mass}{Area\times Thickness}$

$\sf:\implies\:19.30=\dfrac{Mass}{(0.5\times 10^4)\times 0.005}$

$\sf:\implies\:19.30=\dfrac{Mass}{25}$

$\sf:\implies\:Mass=19.30\times 25$

$:\implies\:\underline{\boxed{\bf{\red{Mass=482.5\:g}}}}$

B] Number of moles :

❒ We know that, number of moles is defined as the ratio of mass of element/compound to the atomic/molar mass.

• Molar mass of gold = 196.97

$\sf:\implies\:n=\dfrac{Mass}{Atomic\:mass}$

$\sf:\implies\:n=\dfrac{482.5}{196.97}$

$:\implies\:\underline{\boxed{\bf{\blue{n=2.45}}}}$

C] Number of gold atoms :

❒ We know that, one mole of element contains $\sf6.022\times10^{23}$ atoms.

$\sf:\implies\:N_{atoms}=n\times 6.022\times 10^{23}$

$\sf:\implies\:N_{atoms}=2.45\times (6.022\times 10^{23})$

$\sf:\implies\:N_{atoms}=14.75\times10^{23}$

$\sf:\implies\:\underline{\boxed{\bf{\gray{N_{atoms}=1.48\times10^{24}\:atoms}}}}$

D] At STP, one mole of any gas occupies volume equal to 22.4 litres.