Question

Calculate the volume of 6.022×10²⁰ molecules of Argon gas​.​

Answer 1

$\huge\underbrace\mathfrak\pink{Answer}$

The volume occupied by the 1mole of N

2

=22.4l

So, the 1mole=6.022×10

23

molecules of N

2

=

6.022×10

23

22.4

So, 6.022×10

22

molecules of N

2

would occupy volume

=

6.022×10

23

l

22.4

×6.022×10

22

=2.24l

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Answer 2

Given :

• 6.022 × 10²⁰ molecules of Argon gas

To find :

• Volume of Argon gas

Solution :

☯ Here we have :

⇒ 6.022 × 10²⁰ molecules of Argon gas

☯ Now number of moIes of Argon :

⇒ n = 6.022 × 10²⁰ × (1/6.022 × 10²³)

⇒ n = 10⁻⁰³

☯ Now we know that :

1 mole of any ideal gas occupies 22.4 l

☯ Now we know the foIIowings :

⇒ n = V/22.4

⇒ V = n × 22.4

⇒ V = 10⁻⁰³ × 22.4

⇒ V = 0.0224 L

Therefore, voIume of Argone gas = 0.224 litres.