Respuesta :
Answer:
1.33 kilograms of this liquid must be evaporated to freeze a tray of water at 0°C to ice at 0°C.
Explanation:
Let the mass of liquefied dichlorodifluoromethane be x.
Mass of water to freeze = 571 g
Moles of water =[tex]\frac{571 g}{18 g/mol}=31.7 mol[/tex]
Heat of fusion of ice = 6.02 kJ/mol
Heat lost when 1 mole of water freeze's = -6.02kJ/mol
Heat lost when 31.7 moles of water freeze's: Q
[tex]Q=-6.02 kJ/mol\times 31.7 mol=-191. kJ[/tex]
Heat required to evaporate x amount of liquefied dichlorodifluoromethane: Q'
Q'= -(Q) = 191. kJ
Moles of liquefied dichlorodifluoromethane =[tex]\frac{x}{121 g/mol}[/tex]
Heat of vaporization of dichlorodifluoromethane = 17.4 kJ/mol
[tex]Q'=17.4 kJ/mol\times \frac{x}{121 g/mol}[/tex]
[tex]191. kJ=17.4 kJ/mol\times \frac{x}{121 g/mol}[/tex]
Solving for x:
x = 1328.2 g = 1.33 kg
1 g = 0.001 kg
1.33 kilograms of this liquid must be evaporated to freeze a tray of water at 0°C to ice at 0°C.
