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

##### 1 Answer

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 **by mass** potassium cyanide, i.e.

This means that **for every**

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 **for every** **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

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

#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 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"#