# 0.543g of CaCl2 was dissolved in enough water to make a 500ml solution. What is the concentration of Cl- in %w/v?

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Ok, so I calculated the concentration to be 9.62x10^-3 and then multiplied this by 2 since there are 2 chlorine ions. Using this i used mass=cxvxmm which gave me 0.341g/l. How do i convert this to %w/v? Correct answer is 0.0341%w/v

Ok, so I calculated the concentration to be 9.62x10^-3 and then multiplied this by 2 since there are 2 chlorine ions. Using this i used mass=cxvxmm which gave me 0.341g/l. How do i convert this to %w/v? Correct answer is 0.0341%w/v

##### 1 Answer

Here's what I got.

#### Explanation:

Interestingly enough, I'm not getting

Start by calculating the **percent composition** of chlorine, *chloride anions*,

To do that, use the **molar mass** of calcium chloride, the **molar mass** of elemental chlorine, and the fact that **mole** of calcium chloride contains **moles** of chlorine atoms.

#(2 xx 35.453 color(red)(cancel(color(black)("g mol"^(-1)))))/(110.98 color(red)(cancel(color(black)("g mol"^(-1))))) * 100% = "63.89% Cl"#

This means that for every

As you know, the mass of an *ion* is **approximately** equal to the mass of the neutral atom, so you can say that for every

This implies that your sample contains

#0.543 color(red)(cancel(color(black)("g CaCl"_2))) * "63.89 g Cl"^(-)/(100color(red)(cancel(color(black)("g CaCl"_2)))) = "0.3469 g Cl"^(-)#

Now, in order to find the **mass by volume percent concentration** of chloride anions in the resulting solution, you must determine the mass of chloride anions present in

Since you know that

#100 color(red)(cancel(color(black)("mL solution"))) * "0.3469 g Cl"^(-)/(500color(red)(cancel(color(black)("mL solution")))) = "0.06938 g Cl"^(-)#

Therefore, you can say that the mass by volume percent concentration of chloride anions will be

#color(darkgreen)(ul(color(black)("% m/v = 0.069% Cl"^(-))))#

I'll leave the answer rounded to two **sig figs**, but keep in mind that you have one significant figure for the volume of the solution.

**ALTERNATIVE APPROACH**

Alternatively, you can start by calculating the number of *moles* of calcium chloride present in your sample

#0.543 color(red)(cancel(color(black)("g"))) * "1 mole CaCl"_2/(110.98color(red)(cancel(color(black)("g")))) = "0.004893 moles CaCl"_2#

To find the **molarity** of this solution, calculate the number of moles of calcium chloride present in **moles** present in

#10^3 color(red)(cancel(color(black)("mL solution"))) * "0.004893 moles CaCl"_2/(500color(red)(cancel(color(black)("mL solution")))) = "0.009786 moles CaCl"_2#

You can thus say your solution has

#["CaCl"_2] = "0.009786 mol L"^(-1)#

Since every mole of calcium chloride delivers **moles** of chloride anions to the solution, you can say that you have

#["Cl"^(-)] = 2 * "0.009786 mol L"^(-1)#

#["Cl"^(-)] = "0.01957 mol L"^(-)#

This implies that

#100 color(red)(cancel(color(black)("mL solution"))) * "0.01957 moles Cl"^(-)/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.001957 moles Cl"^(-)#

Finally, to convert this to grams, use the **molar mass** of elemental chlorine

#0.001957 color(red)(cancel(color(black)("moles Cl"^(-)))) * "35.453 g"/(1color(red)(cancel(color(black)("mole Cl"^(-))))) = "0.06938 g Cl"^(-)#

Once again, you have

#color(darkgreen)(ul(color(black)("% m/v = 0.069% Cl"^(-))))#

In reference to the explanation you provided, you have

#"0.341 g L"^(-1) = "0.0341 g/100 mL" = "0.0341% m/v"#

because you have

However, this solution does not contain

#["Cl"^(-)] = "0.01957 mol L"^(-1)#

you have

#n = c * V#

so

#n = "0.01957 mol" * 10^(-3) color(red)(cancel(color(black)("mL"^(-1)))) * 500 color(red)(cancel(color(black)("mL")))#

#n = "0.009785 moles"#

This is how many moles of chloride anions you have in

#100 color(red)(cancel(color(black)("mL solution"))) * "0.009785 moles Cl"^(-)/(500color(red)(cancel(color(black)("mL solution")))) = "0.001957 moles Cl"^(-)#

So once again, you have