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What is the new freezing point of a solution containing 22.5 g NaCl in 375.0 mL of water? (1 mol NaCl = 58.44 g, Kf for water = 1.86°C/m)?

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  • 1 month ago
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    Supposing complete ionization:

    NaCl → Na{+} + Cl{-}  [two ions total]

    (22.5 g NaCl) / (58.44 g NaCl/mol) x (2 mol ions / 1 mol NaCl) / (0.3750 kg) =

    2.0534 mol ions / kg

    (1.86°C/m) x (2.0534 m) = 3.8193°C change

    0°C - 3.8193°C = - 3.82°C

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