Question

Question #6a273

Answer 1

Here's what I got.

Explanation:

FULL QUESTION

A student takes 7.00 g of potassium permanganate, `"KMnO"_text(4(s])`, and dissolves it into 30.0 mL of water. (skipping this question)

a) Calculate the concentration of the resulting solution in mol/L
(3 marks)

b) Calculate the concentration of the resulting solution in parts per million (ppm). Is this an appropriate way to communicate the concentration in this situation ? Support your answer. (4 marks)

c) If the student pours the original solution into a jar that contains 250 mL of water, find the concentration of the resulting solution in the jar in mol/L. (3 marks)

d) If the student pours the original solute into a reservoir that contains `4.00 xx 10^6"g"` of water, calculate the concentration of the resulting solution in ppm. is this an appropriate way to communicate the concentration in this situation ? Support your answer. (4 marks)

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So, you know that you dissolve 7.00 g of potassium permanganate in 30.0 mL of water.

To get the molarity of the resulting solution, which will be defined as the ratio between the number of moles of potassium permanganate and the volume of the solution - in liters.

Now, I think that you're supposed to assume that adding that much potassium permanganate to 30.0 mL of water will not change the volume of the solution.

Let's assume that this is the case. Use potassium permanganate's molar mass to find the number of moles you have

`7.00color(red)(cancel(color(black)("g"))) * "1 mole"/(158.034color(red)(cancel(color(black)("g")))) = "0.04429 moles"`

This means that the molarity of the first solution will be

`C = n/V`

`C = "0.04429 moles"/(30.0 * 10^(-3)"L") = color(green)("1.48 M")`

To get the concentration in parts per million, or ppm, you need to know the mass of water.

Again, I assume that you're supposed to use water's density as being equal to `"1 g/mL"` - this would make the calculations easier.

`30.0color(red)(cancel(color(black)("mL"))) * "1 g"/(1color(red)(cancel(color(black)("mL")))) = "30.0 g"`

Concentration in parts per million can be found by dividing the mass of the solute in grams by the mass of the solvent in grams, and then multiplying the result by `10^6`.

In your case, this will be

`(7.00color(red)(cancel(color(black)("g"))))/(30.0color(red)(cancel(color(black)("g")))) * 10^6 = 2.33 * 10^5"ppm"`

This is not an appropriate way to communicate the concentration of potassium permanganate in the solution because ppm is used to express very small quantities of a substance.

In your case, potassium permanganate is not in trace amounts in solution, so you don't need to use ppm, which denotes parts of solute per million parts of solvent, to express this concentration.

Now, when you add 250 mL of water to this solution, you're essentially diluting it.

That means that the number of moles of the potassium permanganate will remain constant. The only that changes when you dilute a solution is the volume of said solution.

As a result youcan expect the second solution's molarity to decrease. The total volume of this second solution will be

`V_"total" = V_"sol 1" + V_"added water"`

`V_"total" = "30.0 mL" + "250 mL" = "280 mL"`

The molarity of the solution will thus be

`C = "0.04429 moles"/(280 * 10^(-3)"L") = color(green)("0.158 M")`

If you pour the original solution into a reservoir that contains `4.00 * 10^6` grams of water, you an essentially assume that the final mass of the water will be `4.00 * 10^6` grams.

The concentration in ppm will now be

`(7.00color(red)(cancel(color(black)("g"))))/(4.00 * cancel(10^6)color(red)(cancel(color(black)("g")))) * cancel(10^6) = color(green)("1.75 ppm")`

This time you can use ppm to express the concentration of the potassium permanganate because it is now found in very, very small amounts in the solution.

SIDE NOTE Some important assumptions needed to be made here.

I personally do not think that you can dissolve that much potassium permanganate in 30.0 mL of water without increasing the volume of the solution.

Moreover, I don't think that you an do so at a temperature that will allow water's density to be approximated to 1 g/mL.