# 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

Sep 2, 2017

Here's what I got.

#### Explanation:

Interestingly enough, I'm not getting $\text{0.0341% w/v}$ either. Here's why.

Start by calculating the percent composition of chlorine, $\text{Cl}$, in calcium chloride, This will help you calculate the mass of chloride anions, ${\text{Cl}}^{-}$, 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 $\text{100 g}$ of calcium chloride, you get $\text{63.89 g}$ 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 $\text{100 g}$ of calcium chloride, you get $\text{63.89 g}$ of chloride anions, ${\text{Cl}}^{-}$.

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 $\text{100 mL}$ of this solution.

Since you know that $\text{500 mL}$ of solution contain $\text{0.3469 g}$ of chloride anions, you can say that $\text{100 mL}$ 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

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{% m/v = 0.069% Cl}}^{-}}}}$

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.

$\textcolor{w h i t e}{.}$
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 $\text{1 L} = {10}^{3}$ $\text{mL}$ of solution by using the fact that you have $0.004893$ moles present in $\text{500 mL}$ 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 $\text{100 mL}$ 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

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{% m/v = 0.069% Cl}}^{-}}}}$

In reference to the explanation you provided, you have

$\text{0.341 g L"^(-1) = "0.0341 g/100 mL" = "0.0341% m/v}$

because you have $\text{1 L} = {10}^{3}$ $\text{mL}$.

However, this solution does not contain $\text{0.341 g}$ of chloride anions in $\text{1 L}$. Using

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

you have

$n = c \cdot V$

so

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

$n = \text{0.009785 moles}$

This is how many moles of chloride anions you have in $\text{500 mL}$ of solution. Consequently, $\text{100 mL}$ 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 $\text{0.06938 g}$ of chloride anions in $\text{100 mL}$ of solution, the equivalent of $\text{0.069% m/v}$.