calculate the no of ions of AlF3 in two moles?
Answers
Answer:
A 0.21 M solution of AlF contains 0.21 moles of molecules for each liter of solution. In this problem we have only 65.5 mL (0.0655 L) of solution, so the number of moles is
0.21
M
×
0.0655
L
=
0.01376
m
o
l
.
Ionization of each AlF unit produces one fluoride ion (
F
−
), so the number of moles of fluoride ions is the same,
0.01376
m
o
l
. As the other contributors have pointed out, the correct formula for aluminum fluoride is
A
l
F
3
, and in this case, each formula unit would produce three
F
−
ions.
Finally, to obtain the actual number of ions, we multiply by Avogadro's number:
0.01376
m
o
l
×
6.022
×
10
23
m
o
l
−
1
=
8.28
×
10
21
or, in the case of
A
l
F
3
,
3
⋅
0.01376
m
o
l
×
6.022
×
10
23
m
o
l
−
1
=
2.48
×
10
22
Answer:
1 mole of AlF3 contain 1 mole of Al3+ ions and 3 moles F- ions
AlF3 = Al3+ + 3F-
Molar mass of AlF3 = 27 + 3x19 = 84g/mol
⇒ 84g of AlF3 has 3 moles of F-
2.1g of AlF3 will contain = (2.1x3)/84
= 0.075 moles of F-
The Avogadro's number = 6.022 x 10^23
1 mole of F- has 6.022 x 10^23 ion
What about 0.075 moles = 0.075 x 6.022x10^23
= 4.5165 x 10^22
Explanation: