Answer is: 0,0030 mol of carbon dioxide.
Carbon dioxide solubility in water at 20°C and 1 atm is: 3,8·10⁻² mol/L.
Carbon dioxide solubility in water at 25°C and 1 atm is: 3,5·10⁻² mol/L.
Difference is: 0,038 mol/L - 0,035 mol/L = 0,003 mol/L.
V(carbon dioxide) = 1 L.
n(carbon dioxide) = V · c.
n(carbon dioxide) = 1 L · 0,003 mo/L.
n(carbon dioxide) = 0,0030 mol.
