Chemistry help?
2020-08-30 2:52 am
3.52 grams of a water soluble ionic compound with the formula of A1X4 (Molar mass of 125.00 grams/mol) is dissolved in 66 grams of water. Calculate the expected boiling point of the solution. The boiling point elevation constant for water is 0.512 degrees C/molality. The boiling point of water is 98.20 degrees C.
回答 (1)
2020-08-30 5:16 am
Δt = i Kb m
in this case number of moles of compound = 3,52g / 125.00 grams/mol
= 0.02816 moles
these are dissolved in 0.066 kg of water
so m = 0.02816 mole / 0.066 kg = 0.427 m
when AX4 dissolves in water we have 5 particles so i = 5
Δt = 5 * 0.512 * 0.427 = 1.09 degree C
expected boiling point = 1.09 + 98.20 degrees C
= 99.3 degrees C ... answer
in this case number of moles of compound = 3,52g / 125.00 grams/mol
= 0.02816 moles
these are dissolved in 0.066 kg of water
so m = 0.02816 mole / 0.066 kg = 0.427 m
when AX4 dissolves in water we have 5 particles so i = 5
Δt = 5 * 0.512 * 0.427 = 1.09 degree C
expected boiling point = 1.09 + 98.20 degrees C
= 99.3 degrees C ... answer
收錄日期: 2021-04-24 07:57:42
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