Question

67.2L of CO2 gas is filled in the cylinder at STP a) Calculate the number of molecules present in it? b) Calculate the mass of CO2 present in it  Atomic mass  C = 12, O = 16l​

Answer 1

Answer :-

(a) Number of molecules = 1.8066 × 10²⁴

(b) Mass of CO₂ = 132 grams

Explanation :-

We have :-

→ Volume of CO₂ at STP = 67.2 L

→ Avogadro Number = 6.022 × 10²³

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We know that, 1 mole of any gas occupies a volume of 22.4 L at STP. So, number of moles of CO2 gas present are :-

= 67.2/22.4

= 3 moles

Also, molar mass of CO₂ is :-

= 12 + 16(2)

= 12 + 32

= 44 g/mol

Number of molecules :-

= No of moles × Avogadro Number

= 3 × 6.022 × 10²³

= 18.066 × 10²³

= 1.8066 × 10²⁴

Mass of CO₂ :-

= No of moles × Molar mass

= 3 × 44

= 132 g

Answer 2

Answer:

Given :-

• 67.2 L of CO₂ gas is filled in the cylinder at STP.
• Atomic mass (C = 12 , O = 16)

To Find :-

• Calculate the number of molecules present in it.
• Calculate the mass of CO₂ present in it.

Solution :-

First, we have to find the numbers of moles are present in CO₂ gas :

As we know that :

✯︎ 1 moles of gas contains 22.4 L of volume at STP ✯︎

Then,

↦ Numbers of Moles = 67.2/22.4

↦ Numbers of Moles = 3/1

➦ Numbers of Moles = 3 mol

Now, we have to find the molar mass of CO₂ :

Given :

• C = 12
• O = 16

⇒ Molar Mass = CO₂

⇒ Molar Mass = 12 + 16(2)

⇒ Molar Mass = 12 + 16 × 2

➣ Molar Mass = 44 g/mol

Again,

★ The Numbers Of Molecules Present In CO₂ :

As we know that :

✰ Number Of Molecules = Number Of Moles × Avogadro Number ✰

Here,

• Avogadro Number = 6.022 × 10²³

Given :

• Number of Moles = 3 mol
• Avogadro Number = 6.022 × 10²³

According to the question by using the formula we get,

➵ Number Of Molecules = 3 × 6.022 × 10²³

➵ Number Of Molecules = 18.066 × 10²³

➵ Number Of Molecules = 1.8066 × 10²⁴

➲ Number Of Molecules = 1.8066 × 10²⁴

∴ The numbers of molecules present in CO₂ is 1.8066 × 10²⁴ molecules.

Now,

✭ The Mass Present In CO₂ :

As we know that :

✰ Number Of Moles = Mass/Molar Mass ✰

Given :

• Numbers of Moles = 3 mol
• Molar Mass = 44 g/mol

According to the question by using the formula we get,

➟ 3 = Mass/44

By doing cross multiplication we get,

➟ Mass = 44(3)

➟ Mass = 44 × 3

➠ Mass = 132 g

∴ The mass present in CO₂ is 132 g .