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