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

Apr 3, 2018

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. $\text{m/m %}$.

This means that for every $\text{100 g}$ of this solution, you get $\text{0.337 g}$ of potassium cyanide, the solute.

In essence, something like this

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

is supposed to mean

$\textcolor{b l u e}{\text{0.337 g solute") quad "for every" quad color(darkorange)("100 g of the solution}}$

So instead of writing all that information, we use the % sign to mean for every $\text{100 g}$ of the solution.

Now, you know that

$\text{1 g" = 10^3 quad "mg}$

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

$\text{0.337 g KCN" = 0.337 color(red)(cancel(color(black)("g"))) * (10^3 quad "mg")/(1color(red)(cancel(color(black)("g")))) = 0.337 * 10^3 quad "mg KCN}$

and the mass of the solution as

$\text{100 g solution" = 100 color(red)(cancel(color(black)("g"))) * (10^3 quad "mg")/(1color(red)(cancel(color(black)("g")))) = 100 * 10^3 quad "mg solution}$

So instead of saying that your solution contains $\text{0.337 g}$ of potassium cyanide for every $\text{100 g}$ of the solution, you can say that it contains $0.337 \cdot {10}^{3}$ $\text{mg}$ of potassium cyanide for every $100 \cdot {10}^{3} \quad \text{mg}$ of the solution.

In other words, you have

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

as

$\textcolor{b l u e}{0.337 \cdot {10}^{3} \quad \text{mg solute") quad "for every" quad color(darkorange)(100 * 10^3 quad "mg of the solution}}$

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

$\textcolor{b l u e}{\text{0.337 mg solute") quad "for every" quad color(darkorange)("100 mg of the solution}}$

This means that in order for your solution to contain $\text{676 mg}$ 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

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{mass of solution = 201,000 mg}}}}$

To express this in grams, use the conversion factor

$\text{1 g" = 10^3 quad "mg}$

to get

$\text{201,000" color(red)(cancel(color(black)("mg"))) * "1 g"/(10^3color(red)(cancel(color(black)("mg")))) = "201 g}$