A particularrefrigerator cools by evaporating liquefied chlorodifluoromethane,CC

Question

A particularrefrigerator cools by evaporating liquefied chlorodifluoromethane,CCL2F2. How many kilograms of this liquidmust be evaporated to freeze a tray of water at 0' C toice at 0' C? The mass of the water is 525g, the heat of fusion ofice in 6.01 kJ/mol, and the heat of vaporization ofdichlorodifluromethane is 17.4 kJ/mol. Mass of water = 525 g Moles of water = Mass / molar mass =525 g / 18.02 g/mol = 29.13 mol To freeze 29.13 moles of water , heat to be absorbed is 6.01kJ/ mol * 29.13 mol =175.07 kJ Heat of vaporization of dichlorodifluoromethane = 17.4kJ/mol So 1 mole of CCl2F2 absorbs 17.4 kJ Moles of CCl2F2 required = 175.07 kJ * ( 1mole / 17.4 kJ) =10.06 moles Mass of CCl2F2 required = Moles * molar mass = 10.06 mol * 120.91 g/mol =1216.35 g or 1.216 kg

***** I understand how to do this problem, but my question is after you get 10.06 mol,**** What I don't understand is where does the 120.91 g/mol come from to get 1216.35 g 10.06 mol * 120.91 g/mol =1216.35 g or 1.216 kg

Explanation / Answer

Molar mass of CCl2F2 = 120.91 g/mol

literature value for molar mass of CCl2F2 is 120.91 g/mol

moles = grams of solute/molar mass of solute

So,

mass of CCl2F2 liquid required = moles x molar mass of CCl2F2

                                                  = 10.16 mol x 120.91 g/mol

                                                   = 1216.35 g/1000

                                                   = 1.216 kg   

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