# 200 mL of calcium chloride solution contains 3.011 × 1022 Cl ions. Calculate the molarity of the solution. Assume that calcium chloride is completely ionized.

Your friend has correctly answered the question, but has done wrong calculation at the end. Answer will be 0.125 mol/L.

Balanced equation of ionization of CaCl2 will be:

CaCl​2 $\to$ Ca2+ + 2Cl-

No. of moles = ?
Volume = 200 mL = 0.2 L

First we need to calculate the number of moles of solute i.e. CaCl​2.

1 mole of CaCl​2 provides 2 moles of Cl- ions according to our equation.

i.e. $2×6.022×{10}^{23}$ ions of Cl- are produced from = 1 mole of CaCl​2
So, $3.011×{10}^{22}$ ions of Cl​- will be produced from =
=

Thus number of moles of solute = $\frac{1}{40}$

Now,

• 68
Cacl​2------ Ca2++2Cl-

This means 2 mole cl is obtained from 1 mole cacl2
or 2×6.022×1023atoms of cl =1 mole​of cacl2
So 3.011×1022atomsof cl = 3.011×1022/2×6.022×1023 mole​ofcacl2
=
1/40 moles ofcacl2
​Molarity = 1×1000/40×200
=12.5 mol/l
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