Find the freezing point of a solution of 4.4 grams of naphthalene in 100 grams of benzene with Kf = 5.10oC / mol and Mr naphthalene = 128 grams / mol (freezing point of solvent = 5.5oC)
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Answer:
Δt = i Kf m
4.4 °C = (1) (30. °C kg mol¯1) (x / 1.50 kg)
4.4 °C = (1) (20. °C mol¯1) (x)
x = 0.22 mol
0.22 mol times 80.0896 g/mol = 17.6 g (I'll ignore sig figs and leave it at three. I'm such a rebel!)
Explanation:
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