Question

What mass of a 0.337% KCN solution contains 696 mg of KCN? Thank you!

Answer 1

Here's what I got.

Explanation:

Based on the information you've provided, I would say that you're dealing with a potassium cyanide solution that is `0.337%` by mass potassium cyanide, i.e. `"m/m %"`.

This means that for every `"100 g"` of this solution, you get `"0.337 g"` of potassium cyanide, the solute.

In essence, something like this

`color(blue)(0.337) color(darkorange)(%) quad "m/m"`

is supposed to mean

`color(blue)("0.337 g solute") quad "for every" quad color(darkorange)("100 g of the solution")`

So instead of writing all that information, we use the `%` sign to mean for every `"100 g"` of the solution.

Now, you know that

`"1 g" = 10^3 quad "mg"`

This means that you can rewrite the mass of the solute as

`"0.337 g KCN" = 0.337 color(red)(cancel(color(black)("g"))) * (10^3 quad "mg")/(1color(red)(cancel(color(black)("g")))) = 0.337 * 10^3 quad "mg KCN"`

and the mass of the solution as

`"100 g solution" = 100 color(red)(cancel(color(black)("g"))) * (10^3 quad "mg")/(1color(red)(cancel(color(black)("g")))) = 100 * 10^3 quad "mg solution"`

So instead of saying that your solution contains `"0.337 g"` of potassium cyanide for every `"100 g"` of the solution, you can say that it contains `0.337 * 10^3` `"mg"` of potassium cyanide for every `100 * 10^3 quad "mg"` of the solution.

In other words, you have

`color(blue)(0.337) color(darkorange)(%) quad "m/m"`

as

`color(blue)(0.337 * 10^3 quad"mg solute") quad "for every" quad color(darkorange)(100 * 10^3 quad "mg of the solution")`

Finally, you can simplify this by getting rid of the `10^3` to get

`color(blue)("0.337 mg solute") quad "for every" quad color(darkorange)("100 mg of the solution")`

This means that in order for your solution to contain `"676 mg"` of potassium cyanide, it must have a mass of

`676 color(red)(cancel(color(black)("mg KCN"))) * "100 mg solution"/(0.337 color(red)(cancel(color(black)("mg KCN")))) = "200,593.5 mg solution"`

Rounded to three sig figs, the number of sig figs you have for your values, the answer will be

`color(darkgreen)(ul(color(black)("mass of solution = 201,000 mg")))`

To express this in grams, use the conversion factor

`"1 g" = 10^3 quad "mg"`

to get

`"201,000" color(red)(cancel(color(black)("mg"))) * "1 g"/(10^3color(red)(cancel(color(black)("mg")))) = "201 g"`