Question

A sample containing 4.10 g of O2 gas has an initial volume of 21.0 L . What is the final volume, in liters, when each of the following changes occurs in the quantity of the gas at constant pressure and temperature?

A) A sample of 0.300 mole of O2 is added to the 4.10 g of O2 in the container. Express the volume in liters to three significant figures.

B). A sample of 2.20 g of O2 is removed from the 4.10 g of O2 in the container. Express the volume in liters to three significant figures.

C). A sample of 2.20 g of O2 is added to the 4.10 g of O2 gas in the container. Express the volume in liters to three significant figures.

Answer 1

A)

0.300 moles of O₂ gas = molar mass of O₂ x 0.300 = 32 x 0.300 = 9.60 g

Total quantity of O₂ gas in the container now = 4.10 + 9.60 = 13.7 g

4.10 g of O₂ gas has an initial volume of 21.0 L

Therefore 13.7 g O₂ gas will occupy at constant pressure and temperature = (21.0 x 13.7)/4.10

                                                                                                                       = 70.2 L

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B)

If 2.20 g O₂ is removed then the remaining quantity of O₂ = (4.10 - 2.20) = 1.90 g

Now,

4.10 g of O₂ gas has an initial volume of 21.0 L

Therefore 1.90 g O₂ gas will occupy at constant pressure and temperature = (21.0 x 1.90)/4.10

                                                                                                                       = 9.73 L

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C)

If 2.20 g O₂ is added then the total quantity O₂ = 4.10 + 2.20 = 6.30 g

Now,

4.10 g of O₂ gas has an initial volume of 21.0 L

Therefore 6.30 g O₂ gas will occupy at constant pressure and temperature = (21.0 x 6.30)/4.10

                                                                                                                       = 32.3 L