Question

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

Answer 1

Answer:

We can use the Ideal Gas Law to solve this problem. ... NH3 ×1 mol NH3 6.022×1023molecules NH3 =0.100 02 mol NH 3.

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.