Aballoon filled with helium has a volume of 43.4 l at 305 k. what volume will the balloon occupy at 273 k?

To solve this, you need to use a relationship for gases known as Charles' Law. Charles' Law says that at a constant pressure, volume and temperature have a direct proportionality relationship. The equation accompanying this is:
(V1)/(T1) = (V2)/(T2)
We have 3 known values so we can solve for the fourth:
V1 = 43.4 L
T1 = 305 K
T2 = 273 K
V2 = Unknown
Plugging into our equation:
(43.4 L)/(305 K) = (V2)/(273 K)
V2 = (43.4 L/305K) * (273 K)
V2 = 38.8 L = Volume the balloon occupies at 273 K
This makes sense, as a hotter gas will have more average kinetic energy, and would take up a large volume. When the volume was calculated at a lower temperature, the calculated volume decreased as expected.

The Answer would be e: 3.2 g/L 

At STP : p = 1 atm and T = 273.15 K 
molar mass Cl2 = 70.906 g/mol 

d = molar mass x p/RT = 70.906 x 1 atm / 0.08206 x 273.15=3.16 g/L