Volume of 2.99×10²³ chlorine molecules at STP

[called2serve]

Can someone please calculate the volume occupied at STP by 2.99times10 to the 23rd power molecules of Chlorine. Thank you Very much.

 

[ShawnD]

All gases have a specific size at STP. I think it's 22.4L IIRC. Just multiply 22.4 * (2.99 * 10^23) / (avogadro's number)

 

[lvlastermind]

All gases have a specific size at STP. I think it's 22.4L IIRC. Just multiply 22.4 * (2.99 * 10^23) / (avogadro's number)

avogadro's number = 6.02*10^23

incase you didn't know

 

[Gokul43201]

1. Use the Ideal Gas Law, PV =nRT, with n= $$\frac {2.99*10^{23}} {6.02*10^{23}}$$, to find V. You know, P, T, R, n.

OR

2. At STP, one mole of an ideal gas occupies 22.4 liters. One mole has 6.02*10^23 molecules. Use ratios to find the volume of 2.99*10^23 molecules.