What is the freezing point (in degrees Celcius) of 4.81 kg of water if it contains 154.5 g of CaBr2?

The freezing point depression constant for water is 1.86 degrees celsius/m and the molar mass of CaBr2 is 199.89 g/mol

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  • 1 month ago

    Supposing complete ionization:

    CaBr2 → Ca{2+} + 2 Br{-}   [three ions total]

    (154.5 g CaBr2) / (199.886 g CaBr2/mol) x (3 mol ions / 1 mol CaBr2) / (4.91 kg) =

    0.472265 mol/kg = 0.472265 m ions

    (0.472265 m ions) x (1.86 °C/m) = 0.878°C change

    0.000°C - 0.878°C =  - 0.878°C

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