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200 mL of calcium chloride solution contains 3.011 × 10^{22} Cl^{ } ions. Calculate the molarity of the solution. Assume that calcium chloride is completely ionized.

Balanced equation of ionization of CaCl

_{2}will be:

CaCl

_{2}$\to $ Ca

^{2+}+ 2Cl

^{-}

$Molarity=\frac{No.ofmolesofsolute}{VolumeofsolutioninL}$

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. $2\times 6.022\times {10}^{23}$ ions of Cl

^{-}are produced from = 1 mole of CaCl

_{2}

So, $3.011\times {10}^{22}$ ions of Cl

^{-}will be produced from = $\frac{1moleofCaC{l}_{2}}{2\times 6.022\times {10}^{23}ions}\times 3.011\times {10}^{22}ions$

= $\frac{1}{40}molesofCaC{l}_{2}$

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

Now,

$Molarityofthesolution=\frac{1}{40\times 0.2}\phantom{\rule{0ex}{0ex}}=0.125mol/L$

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