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hichkok12 [17]
3 years ago
6

At –45oC, 71 g of fluorine gas take up 6843 mL of space. What is the pressure of the gas, in kPa?

Chemistry
1 answer:
Anika [276]3 years ago
4 0
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
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