Question

3.6 gram of oxygen is adsorbed on 1.2g of metal powder. What volume of oxygen in litres adsorbed per gram of the adsorbent at STP?

Answer 1

$\bigstar$ Answer $\bigstar$

$\star$ Given:

3.6 gram of $\rm O_{2}$ is adsorbed on 1.2 g of metal powder

$\star$ To Find:

Volume of $\rm O_{2}$ (in litres) adsorbed per gram of the adsorbent at STP

$\star$ Solution:

Given that,

3.6 gram of $\rm O_{2}$ is adsorbed on 1.2 g of metal powder

Therefore,

Mass of $\rm O_{2}$ gas adsorbed per gram of metal powder

$\red{\implies \rm m=\dfrac{3.6}{1.2} \ g}$

$\red{\implies \rm m=3 \ g}$

We know that,

Ideal Gas Equation

$\green{\boxed{\boxed{\rm PV=nRT}}}$

Here,

P = Pressure (Pa)

V = Volume (m³)

n = Number of moles

R = Gas Constant = 0.0821 L.atm/mol.K

T = Temperate (Kelvin)

According to Question,

We are asked to find that, Volume of $\rm O_{2}$ (in litres) adsorbed per gram of the adsorbent at STP

Given that,

To Find that, Volume of $\rm O_{2}$ (in litres) adsorbed per gram of the adsorbent at STP.

STP = Standard Temperature and Pressure

Therefore,

P = 1 atm

R = 0.0821 L.atm/mol.K

T = 273 Kelvin

We know that,

n = Number of moles

It can also be written,

$\rm \implies n=\dfrac{m}{M}$

We found out,

m = 3 g

And, Gram Molecular Weight of $\rm O_{2}$

M = 32 g

So the modified formula,

$\blue{\boxed{\boxed{\rm PV=\dfrac{m}{M}RT}}}$

According to Question,

We are asked to find that, Volume of $\rm O_{2}$ (in litres) adsorbed per gram of the adsorbent at STP

So,

Volume (V)

$\green{\boxed{\boxed{\rm V=\dfrac{m \times R \times T}{M \times P}}}}$

Here,

P = 1 atm

R = 0.0821 L.atm/mol.K

T = 273 K

m = 3 g

M = 32 g

Substituting the above values,

We get,

$\blue{\rightarrow \rm V=\dfrac{3 \times 0.0821 \times 273}{32 \times 1} \ Lg^{-1}}$

On further Simplification,

We get,

$\pink{\rm \implies V=2.1 \ Lg{-1}}$

Hence,

Volume of $\rm O_{2}$ adsorbed per gram of the adsorbent is 2.1 Lg⁻¹

$\checkmark$ V =  2.1 Lg⁻¹

Answer 2

Answer:

$\bigstar$ Answer $\bigstar$

$\star$ Given:

3.6 gram of $\rm O_{2}$ is adsorbed on 1.2 g of metal powder

$\star$ To Find:

Volume of $\rm O_{2}$ (in litres) adsorbed per gram of the adsorbent at STP

$\star$ Solution:

Given that,

3.6 gram of $\rm O_{2}$ is adsorbed on 1.2 g of metal powder

Therefore,

Mass of $\rm O_{2}$ gas adsorbed per gram of metal powder

$\red{\implies \rm m=\dfrac{3.6}{1.2} \ g}$

$\red{\implies \rm m=3 \ g}$

We know that,

Ideal Gas Equation

$\green{\boxed{\boxed{\rm PV=nRT}}}$

Here,

P = Pressure (Pa)

V = Volume (m³)

n = Number of moles

R = Gas Constant = 0.0821 L.atm/mol.K

T = Temperate (Kelvin)

According to Question,

We are asked to find that, Volume of $\rm O_{2}$ (in litres) adsorbed per gram of the adsorbent at STP

Given that,

To Find that, Volume of $\rm O_{2}$ (in litres) adsorbed per gram of the adsorbent at STP.

STP = Standard Temperature and Pressure

Therefore,

P = 1 atm

R = 0.0821 L.atm/mol.K

T = 273 Kelvin

We know that,

n = Number of moles

It can also be written,

$\rm \implies n=\dfrac{m}{M}$

We found out,

m = 3 g

And, Gram Molecular Weight of $\rm O_{2}$

M = 32 g

So the modified formula,

$\blue{\boxed{\boxed{\rm PV=\dfrac{m}{M}RT}}}$

According to Question,

We are asked to find that, Volume of $\rm O_{2}$ (in litres) adsorbed per gram of the adsorbent at STP

So,

Volume (V)

$\green{\boxed{\boxed{\rm V=\dfrac{m \times R \times T}{M \times P}}}}$

Here,

P = 1 atm

R = 0.0821 L.atm/mol.K

T = 273 K

m = 3 g

M = 32 g

Substituting the above values,

We get,

$\blue{\rightarrow \rm V=\dfrac{3 \times 0.0821 \times 273}{32 \times 1} \ Lg^{-1}}$

On further Simplification,

We get,

$\pink{\rm \implies V=2.1 \ Lg{-1}}$

Hence,

Volume of $\rm O_{2}$ adsorbed per gram of the adsorbent is 2.1 Lg⁻¹

$\checkmark$ V =  2.1 Lg⁻¹