Question

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

Answer 1

Here's what I got.

Explanation:

Interestingly enough, I'm not getting `"0.0341% w/v"` either. Here's why.

Start by calculating the percent composition of chlorine, `"Cl"`, in calcium chloride, This will help you calculate the mass of chloride anions, `"Cl"^(-)`, present in your sample.

To do that, use the molar mass of calcium chloride, the molar mass of elemental chlorine, and the fact that `1` mole of calcium chloride contains `2` 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 `"100 g"` of calcium chloride, you get `"63.89 g"` of chlorine.

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 `"100 g"` of calcium chloride, you get `"63.89 g"` of chloride anions, `"Cl"^(-)`.

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 `"100 mL"` of this solution.

Since you know that `"500 mL"` of solution contain `"0.3469 g"` of chloride anions, you can say that `"100 mL"` of solution will contain

`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.

`color(white)(.)`
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 `"1 L" = 10^3` `"mL"` of solution by using the fact that you have `0.004893` moles present in `"500 mL"` of solution.

`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 `2` 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 mL"` of this solution will contain

`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 `"1 L" = 10^3` `"mL"`.

However, this solution does not contain `"0.341 g"` of chloride anions in `"1 L"`. Using

`["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 `"500 mL"` of solution. Consequently, `"100 mL"` of solution will contain

`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 `"0.06938 g"` of chloride anions in `"100 mL"` of solution, the equivalent of `"0.069% m/v"`.