If all ionic substances completely ionized, the aqueous solution with the lowest freezing point would be
1)0.25m NH4NO3
2)0.30m BaCl2
3) 0.35m LiBr
4) 0.50m Sucrose
5) 0.40m MgSO4
I know that i have to multiply m and i to find the largest delta T. I am just not sure if the values that I got for i are correct.
I know that sucrose is 1.
NH4NO3 = 4
BaCl2 = 3 Are these correct?
LiBr = 2
MgSO4 = 3
No.
ammonium nitrate breaks into two ions, NH4+, and NO3-
magnesium sulfate breaks into two ions, Mg++, and SO4--
so NH4NO3 is 2 and MgS04 is 4?
No, MgSO4 is 2. Look at Bob Pursley's post. You get Mg++ and SO4--. 1 + 1 = 2.
Ok thank you!
so 0.30m BaCl2 has the largest delta T at 0.90 so it has the lowest freezing point. Would you agree?
I agree.
To determine the correct values for the ionization factor (i), you need to consider the number of ions that each compound dissociates into when it completely ionizes in solution.
For NH4NO3, it dissociates into NH4+ and NO3- ions. Therefore, the ionization factor (i) for NH4NO3 is 2.
For BaCl2, it dissociates into Ba2+ and 2 Cl- ions. Therefore, the ionization factor (i) for BaCl2 is 3.
For LiBr, it dissociates into Li+ and Br- ions. Therefore, the ionization factor (i) for LiBr is 2.
For MgSO4, it dissociates into Mg2+ and SO42- ions. Therefore, the ionization factor (i) for MgSO4 is 3.
Using the correct values of the ionization factor (i) for each compound, you can now calculate the largest delta T.
To do this, multiply the molality (m) with the ionization factor (i) for each compound and compare the results. The compound with the largest resulting value will have the lowest freezing point.
For example, to find the largest delta T for 0.25m NH4NO3:
Delta T = m * i = 0.25 * 2 = 0.5
Similarly, you can calculate the delta T for the other compounds and compare them to find the compound with the largest delta T, which corresponds to the lowest freezing point.