What is the volume occupied by 3.0*10^233.01023 molecules of bromine gas at STP?

1 Answer
Jul 19, 2016

color(magenta)("11 L")11 L

Explanation:

color(orange)("Because we are at STP, we will have to use the ideal gas law:")Because we are at STP, we will have to use the ideal gas law:
slideplayer.comslideplayer.com

At standard temperature and pressure, the temperature is 273K and the pressure is 1 atm.

Next, list your known and unknown variables. Our only unknown is the volume of Br_2(g)Br2(g). Our known variables are P,R, and T.

We don't necessarily have nn yet, but we're given a value that will lead us to the number of moles of bromine gas. All we have to do is convert molecules to moles using the conversion factor below:www.slideshare.netwww.slideshare.net

We'll use the second conversion factor because we can cancel out molecules and end up with moles:

3.0xx10^(23)cancel"molecules"xx(1mol)/(6.02x10^(23)cancel"molecules") = 0.498 mol

Now all of the variables have good units! All that's left to do is rearrange the equation and solve for V like so:

V = (nxxRxxT)/P
V = (0.498cancel"mol"xx0.0821(Lxxcancel"atm")/(cancel"mol"xxcancel"K")xx(273cancelK))/(1cancel"atm")
V = 11 L