Question

number of atoms present in 2.8litres of diatomic gas at STP

Answer 1

2.8*6.022*10raise to power 23

Answer 2

Answer : The number of atoms present in 2.8 L is, $15.056\times 10^{22}$.

Solution : Given,

Volume of diatomic gas = 2.8 L

At STP,

22.4 L volume of gas contains $2\times 6.023\times 10^{23}$ atoms

2.8 L volume of gas contains $2\times 6.023\times 10^{23}\times \frac{2.8L}{22.4L}=15.056\times 10^{22}$ atoms

Therefore, the number of atoms present in 2.8 L is, $15.056\times 10^{22}$