# Chemistry help please? Assume the solubility of carbon dioxide gas in pop at 5 °C is 0.586 g/100 mL and at 20 °C its solubilty is 0.169 g/100 mL. What volume of carbon dioxide gas will escape...

Assume the solubility of carbon dioxide gas in pop at 5 °C is 0.586 g/100 mL and at 20 °C its solubilty is 0.169 g/100 mL. What volume of carbon dioxide gas will escape from a 710 mL bottle of Pepsi that has been taken out of the fridge and has been sitting open at 20 °C and 102 kPa?
ndnordic | Certified Educator

First you need to know the amount of CO2 released in grams.

At 5 degrees, the total mass of CO2 is 0.586g/100 * 710 ml = 4.16 g of CO2.

At 20 degrees, the total mass of CO2 in the soda is 0.169g/100mL * 710 mL = 1.2 g CO2.

So 4.16-1.2 = 2.96 g of CO2 was released.

Now, one mole of  a gas has a  volume of 22.4 L at STP.

2.96 g/44 g/mole = 0.067 moles of CO2 released.

0.067 moles * 22.4 L/mole = 1.507 L of CO2 at STP.

But you are not at STP!  So use gas laws to convert the volume:

1.507 * (293.1/273.1)*(101.32/102) = 1.607 L of gas at 20 degrees C and 102 kPa.