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
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
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"#