# How do I calculate the number of chloride ions in a solution?

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Which of the following contains the most chloride ions?

A) #10cm^3# of #3.3xx10^-2 mol dm^-3# aluminium chloride solution

B) #20cm^3# of #5.0xx10^-2 mol dm^-3 # calcium chloride solution

C)#30cm^3# of #3.3xx10^-2 mol dm^-3 # hydrochloric acid

D)#40cm^3# of #2.5xx10^-2 mol dm^-3 # sodium chloride solution

Which of the following contains the most chloride ions?

A)

B)

C)

D)

##### 1 Answer

Here's how you can do that.

#### Explanation:

Your strategy here will be to use the molarity and volume of the solutions to figure out how many moles of salt were *dissolved* to make the solution.

Once you know that, you can use the chemical formula of the salt to figure out how many moles of chloride anions,

I'll show you how to solve points **A)** and **B)** and leave the rest to you as practice.

So, **molarity** is defined as the number of moles of solute present in

#10 color(red)(cancel(color(black)("cm"^3))) * "1 dm"^3/(10^3color(red)(cancel(color(black)("cm"^3)))) = 10^(-2)color(white)(.)"dm"^3#

This means that the first solution will contain

#10^(-2) color(red)(cancel(color(black)("dm"^3))) * (3.3 * 10^(-2)color(white)(.)"moles AlCl"_3)/(1color(red)(cancel(color(black)("dm"^3)))) = 3.3 * 10^(-4)color(white)(.)"moles AlCl"_3#

Now, **every mole** of aluminium chloride that dissociates in aqueous solution produces

,one moleof aluminium cations#1 xx "Al"^(3+)# ,three molesof chloride anions#3 xx "Cl"^(-)#

This means that **for every** **mole** of aluminium chloride dissolved to make this solution, you will end up with

#3.3 * 10^(-2) color(red)(cancel(color(black)("moles AlCl"_3))) * "3 moles Cl"^(-)/(1color(red)(cancel(color(black)("mole AlCl"_3)))) = 9.9 * 10^(-4)color(white)(.)"moles Cl"^(-)#

Now do the same for the calcium chloride solution. You will have

#20 color(red)(cancel(color(black)("cm"^3))) * "1 dm"^3/(10^3color(red)(cancel(color(black)("cm"^3)))) = 2.0 * 10^(-2)color(white)(.)"dm"^3#

This means that the solution contains

#2.0 * 10^(-2)color(red)(cancel(color(black)("dm"^3))) * (5.0 * 10^(-2)color(white)(.)"moles CaCl"_2)/(1color(red)(cancel(color(black)("dm"^3)))) = 1.0 * 10^(-3)color(white)(.)"moles CaCl"_2#

Now, **every mole** of calcium chloride that dissociates in aqueous solution will produce

,one moleof calcium actions#1 xx "Ca"^(2+)# ,two molesof chloride anions#2 xx "Cl"^(-)#

This means that the second solution will contain

#1.0 * 10^(-3)color(red)(cancel(color(black)("moles CaCl"_2))) * "2 moles Cl"^(-)/(1color(red)(cancel(color(black)("mole CaCl"_2)))) = 2.0 * 10^(-3)color(white)(.)"moles Cl"^(-)#

Therefore, you can say the solution **B)** will contain the higher concentration of chloride anions, since

#2.0 * 10^(-3)color(white)(.)"moles Cl"^(-) > 9.9 * 10^(-4)color(white)(.)"moles Cl"^(-)#

Use the same approach to calculate the number of moles of chloride anions present in solutions **C)** and **D)** and compare all four solutions to find the answer.