Question

How many grams of ACl3 are required to make 333mL of solution with 3.00M chloride ions?

A 14.8g

B.44.4 g

C. 133g

D. 400g

E 1.20 kg

Answer 1

[Cl-] = 3.00 M

1 mol of AlCl3 has 3 Cl- ion

So,
[AlCl3] = [Cl-]/3
= 3.00 M / 3
= 1.00 M

volume , V = 3.33*10^2 mL
= 0.333 L


use:
number of mol,
n = Molarity * Volume
= 1*0.333
= 0.333 mol

Molar mass of AlCl3,
MM = 1*MM(Al) + 3*MM(Cl)
= 1*26.98 + 3*35.45
= 133.33 g/mol

use:
mass of AlCl3,
m = number of mol * molar mass
= 0.333 mol * 1.333*10^2 g/mol
= 44.4 g
Answer: 44.4 g