In a study of the properties of an aquoous solution of Th(NO3)4 a freezing point depression of 0.0703K was observed for an aqueous solution of molality 9.6 mmol kg^-1 What is the apparent number of ions per formula unit?
2007-11-14 16:33:51 · 5 answers · asked by ShouldBeSleeping 2 in Science & Mathematics ➔ Chemistry
Essentially, you need to find out what the freezing point depression would have been for a 1 molal solution of the material and divide by the molal freezing point depression for water to get apparent # of ions.
[0.0703K/0.0096molal]/1.86K/molal = 3.93 (essentially 4)
2007-11-14 16:50:58 · answer #1 · answered by Flying Dragon 7 · 0⤊ 0⤋
delta T = Kfp(molality) = (1.86 K)(0.0096 m) = 0.0179 K
0.0703 / 0.0179 = 3.93
It appears that the Th(NO3)4 is effectively behaving as though it has four ions per formula unit instead of five as might be expected.
Answer: 4
2007-11-14 16:56:31 · answer #2 · answered by Dennis M 6 · 0⤊ 0⤋
.0096 molal
0.0703 deg K /1.86 = 0.0378 apparent ion molality
0.0378 / 0.0096 = 3.94 apparent ions per formula unit
2007-11-14 16:46:11 · answer #3 · answered by skipper 7 · 0⤊ 0⤋
.32 g of geraniol was lost to the vapor state. How many mols is this?? (n)= Mol = g/MWT You know the the volume of the container and Kelvin temp. Do you remember PV=nRT? Solve for P and plug the numbers in and that will give you the partial pressure for the geraniol.
2016-05-23 05:46:04 · answer #4 · answered by ? 3 · 0⤊ 0⤋
it differs according the different isotopes of the elements you use.
2007-11-14 16:51:41 · answer #5 · answered by Azmi R 2 · 0⤊ 1⤋