Question

10) a. Calculate the solubility in (mol/L) of CO2(g) inside a can containing 355 cm; of water at 25.00°C if the pressure of the CO2(g) inside the can is 2.50 bar. The density of water is 0.9970 (g/cm3) at 25.00°C. b. Now suppose you pop open the can in Flagstaff and the pressure above the solution is reduced to 0.779 bar. How many moles of CO2(g) will eventually be released from the solution into the atmosphere?

Answer 1

We know, according to the ideal gas equation;

PV =nRT

whereas, P = pressure

V = volume

R = gas constant

n = moles

T = Temperature

now put the values, we get;

2.5bar * (355/1000)L = n* 8.314*10^-2 L.bar.K^-1mol^-1 *298K

n = 2.5bar * (355/1000)L / 8.314 *10^-2 L.bar.K^-1mol^-1 * 298K

n = 0.0358 moles

Solubility of CO₂ = Moles of CO₂ dissolved / Volume of solution

= 0.0358moles / 0.335L

= 0.10698moles / L

Now, when the pressure is decreased:

the moles of CO₂ left in solution is:

n = 0.779 bar * (355/1000)L / 8.314 *10^-2 L.bar.K^-1mol^-1 * 298K

= 0.0112 moles of CO₂

So, the moles of CO₂ loss = initial CO₂ moles - left moles of CO₂

the moles of CO₂ loss =  (0.0358 - 0.0112)moles

= 0.0246moles of CO₂

Hence, 0.0246moles of CO₂ is eventually released from the solution.

Hope this is helpful!