Which option contains a #2g# mass of sodium atoms.... #1.# #5.24xx10^22"sodium atoms?"# #2.# #6.022xx10^24"sodium atoms?"# #3.# #3.12xx10^23"sodium atoms?"# #4.# #5.24xx10^22*"sodium atoms?"#

Question

1268 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-08T19:41:14

I would say #"option 1........"#

We know that #1*mol# contains #6.022xx10^23# individual sodium atoms, and has a mass of #22.99*g#.

And thus for #1.# there is a #2*g# mass of #Na#.

For #2.# there is a #22.99*g# mass of #Na#.

For #3.# there is a ..............................

#(3xx10^23*"sodium atoms")/(6.022xx10^23*"sodium atoms"*mol^-1)xx22.99*g*mol^-1# mass of #Na#, approx. #11.5*g#.

For #4.# there is a #8*g# mass of sodium............

And so, clearly,..................Capisce?

Which option contains a #2*g# mass of sodium atoms.... #1.# #5.24xx10^22*"sodium atoms?"# #2.# #6.022xx10^24*"sodium atoms?"# #3.# #3.12xx10^23*"sodium atoms?"# #4.# #5.24xx10^22*"sodium atoms?"#