Question

W p value it needed for this question. Visited In the laboratory you dissolve 22.2 g of aluminum iodide in a volumetric flask and add water to a total volume of 500 mL. What is the molarity of the solution? M. What is the concentration of the aluminum cation? M. What is the concentration of the iodide anion? M. Submit Answer Retry Entire Group 9 more group attempts remaining

Answer 1

1)
a)

Molar mass of AlI3,
MM = 1*MM(Al) + 3*MM(I)
= 1*26.98 + 3*126.9
= 407.68 g/mol


mass(AlI3)= 22.2 g

use:
number of mol of AlI3,
n = mass of AlI3/molar mass of AlI3
=(22.2 g)/(4.077*10^2 g/mol)
= 5.445*10^-2 mol
volume , V = 5*10^2 mL
= 0.5 L


use:
Molarity,
M = number of mol / volume in L
= 5.445*10^-2/0.5
= 0.1089 M
Answer: 0.109 M

b)
1 AlI3 has 1 Al3+
So,
[Al3+] = [AlI3]
= 0.109 M

Answer: 0.109 M

c)
1 AlI3 has 3 I-
So,
[I-] = 3*[AlI3]
= 3*0.109 M
= 0.327 M

Answer: 0.327 M

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