Asked by Jessie
20) The ideal value of i (van't Hoff factor) for (NH4)3PO4.
A) 2 B) 5 C) 4 D) 3 E) 1
I know the answer but how are problems like this figured out?
A) 2 B) 5 C) 4 D) 3 E) 1
I know the answer but how are problems like this figured out?
Answers
Answered by
DrBob222
You look at how they ionize.
(NH4)3PO4 ==> 3NH4^+ + PO4^-3
So that's 4 particles.
NaCl is 2.
CsCl = 2
BaCl2 = 3
Ba3(PO4)2 = 5
C12H22O11(sucrose; i.e., table sugar) = 1 since it dissolves BUT doesn't ionize.
Glucose, C6H12O6 = 1
(NH4)3PO4 ==> 3NH4^+ + PO4^-3
So that's 4 particles.
NaCl is 2.
CsCl = 2
BaCl2 = 3
Ba3(PO4)2 = 5
C12H22O11(sucrose; i.e., table sugar) = 1 since it dissolves BUT doesn't ionize.
Glucose, C6H12O6 = 1
Answered by
Raghav
In these questions, we are not concerned that which compound it is. The only thing which matter is that how they ionise because it is the number of particles that we are concerned about. That's why these are called collimating properties. So 1 molecule of NaCl gives 1 Na atom and 1 Cl atom in water. So, it's vant hot factor (i) = 2.
Answered by
Njabulo
2
Answer
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Answer
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