chem help?
2020-08-31 8:03 am
3.86 grams of a water soluble ionic compound with the formula of A1X2 (Molar mass of 125.00 grams/mol) is dissolved in 161 grams of water. Calculate the expected freezing point of the solution. The freezing point depression constant for water is 1.86 degrees C/molal. The freezing point of water is 0.26 degrees C.
回答 (1)
2020-08-31 8:29 am
moles solute = 3.86 g / 125.00 g/mol = 0.03088 mol
molality of solute = 0.03088 mol / 0.161 kg water = 0.192 molal
Change in freezing point = i Kf m
change in FP = 3 (1.86 C/m) (0.192 m) = 1.07 C.
So, the solution will freeze at 0.26 C - 1.07 C = - 0.81 C
molality of solute = 0.03088 mol / 0.161 kg water = 0.192 molal
Change in freezing point = i Kf m
change in FP = 3 (1.86 C/m) (0.192 m) = 1.07 C.
So, the solution will freeze at 0.26 C - 1.07 C = - 0.81 C
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