At –45oc, 71 g of fluorine gas take up 6843 ml of space. what is the pressure of the gas, in kpa?

The moles of fluorine present are 71/19 = 3.74
Now, we know that one mole of gas at 273 K and 101.3 kPa (S.T.P.) occupies 22.4 liters
Volume of 3.74 moles at S.T.P = 3.74 x 22.4
Volume = 83.776 L = 83,776 mL

Now, we use Boyle's law, that for a given amount of gas,
PV = constant

P x 6843 = 101.3 x 83776
P = 1,240 kPa

Ethene molecular fórmula= c2h4molecular weight = 29.046g/molhope it : )