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 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?
1 Answer | Add Yours
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.
We’ve answered 319,854 questions. We can answer yours, too.Ask a question