Question

Dinitrogen monixide , or laughing gas, is ised as an anesthic. if 0.248 L of laughing gas are inhaled in one breath at STP, how many molecules of dinitrogen monixide are present?

Answer 1

Solution:

At STP,

1 mol of N2O contains 22.4 L = 6.023 x 10^23 molecules

Since, 22.4 L of N2O contains 6.023 x 10^23 molecules

Hence, 0.248 L contains

= (0.248 L x 6.023 x 10^23 molecules) / 22.4 L = 0.0667 x 10^23 molecules

= 6.67 x 10^21 molecules