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

1 Answer
Sep 2, 2017

Answer:

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"#.