What is the freezing point (in C) of a 1.56m aqueous solution of CaCI2? (Report amount to three decimal points.)?
回答 (1)
2016-06-22 5:20 pm
1 mole of CaCl₂ contains 3 moles of ions.
molarity of ions, m = (1.56 m) × 3 = 4.68 m
Molal freezing point depression of water, Kf = 1.86 °C/m
Depression of freezing point = Kf × m = (1.86 °C/m) × (4.68 m) = 8.705°C
Freezing point = (0 - 8.705)°C = -8.705°C
molarity of ions, m = (1.56 m) × 3 = 4.68 m
Molal freezing point depression of water, Kf = 1.86 °C/m
Depression of freezing point = Kf × m = (1.86 °C/m) × (4.68 m) = 8.705°C
Freezing point = (0 - 8.705)°C = -8.705°C
收錄日期: 2021-04-18 15:09:31
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