Answer:
0.5286 kilogram of sulfur dioxide must be evaporated .
Explanation:
Enthalpy of vaporization of chlorofluorocarbons [tex]H_{vap}[/tex]= 20.1 kJ/mol
Mass of chlorofluorocarbons = 1.24 kg = 1240 g
Moles of chlorofluorocarbons = [tex]\frac{1240 g}{121 g/mol}=10.2479 mol[/tex]
Heat removes by 10.2479 moles of chlorofluorocarbons : q
[tex]q = H_{vap}\times moles[/tex]
[tex]q=20.1 kJ/mol \times 10.2479 mol=205.983 kJ[/tex]
Let moles of sulfur dioxide removing 205.983 kJ(q) of heat be n.
Enthalpy of vaporization of sulfur dioxide[tex]H'_{vap} = 5.96 kcal/mol = 24.93664 kJ/mol[/tex]
(1 kcal = 4.184 kJ)
[tex]q=H'_{vap}\times n[/tex]
[tex]205.983 kJ=24.93664 kJ/mol\times n[/tex]
n = 8.2603 moles
Mass of 8.2603 moles of sulfur dioxide =
[tex]8.2602 mol\times 64 g/mol=528.6592 g=0.5286 kg[/tex]
0.5286 kilogram of sulfur dioxide must be evaporated .
