Respuesta :
Answer:
1.22 kilograms of dichlorodifluoromethane must be evaporated to freeze a tray of water at 0 °C to ice at 0 °C.
Explanation:
Mass of water to be freeze = 525 g
Moles of water = [tex]\frac{525 g}{18 g/mol}=29.17 mol[/tex]
Heat of fusion of ice = 6.01 kJ/mol
Heat lost on formation of ice = -6.01 kJ/mol
Heat lost of formation of ice from 29.17 moles of water = q
[tex]q=-6.01 kJ/mol\times 29.17 kJ/mol=-175.297 kJ[/tex]
Heat absorbed by the n moles of liquefied dichlorodifluoromethane will be eqaul to the heat lost by the 29.17 moles of water.
Q = -q = 175.297 kJ/mol
Heat of vaporization of dichlorodifluoromethane is 17.4 kJ mol–1
[tex]175.297 kJ=n\times 17.4 kJ/mol[/tex]
[tex]n=\frac{175.297 kJ}{17.4 kJ/mol}=10.074 mol[/tex]
Mass of 10.074 moles of dichlorodifluoromethane :
10.074 mol × 121 g/mol = 1,218.954 g
1,218.954 g =1,218.954 × 0.001 kg =1.218954 kg ≈ 1.22 kg
1 g = 0.001 kg
1.22 kilograms of dichlorodifluoromethane must be evaporated to freeze a tray of water at 0 °C to ice at 0 °C.
