Question

An aqueous solution of 4.62 M potassium bromide, KBr, has a density of 1.37 g/mL. What is the percent by mass of KBr in the solution?

Answer 1

`"40.1% w/w"`

Explanation:

Your strategy here will be to

• pick a sample volume of this solution and use its density to determine its mass

• use the solution's molarity to find how many moles of potassium bromide it contains

• use potassium bromide's molar mass to determine how many grams of the compound will contain that many moles

Since you're dealing with molarity, which as you know is defined as moles of solute, in this case potassium bromide, `"KBr"`, per liters of solution, you can make the calculations easier by picking a `"1.00-L"` sample of this solution.

This sample will have a mass of

`1.00 color(red)(cancel(color(black)("L"))) * (1000 color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "1.37 g"/(1color(red)(cancel(color(black)("mL")))) = "1370 g"`

Since you know the molarity of this solution, you can say that this `"1.00-L"` sample will contain

`color(blue)(c = n/V implies n = c * V)`

`n_(KBr) = "4.62 M" * "1.00 L" = "4.62 moles KBr"`

Potassium bromide has a molar mass of `"119.0 g/mol"`, which means that one mole of potassium bromide has a mass of `"119 g"`. In your case, `4.62` moles of potassium bromide will have a mass of

`4.62 color(red)(cancel(color(black)("moles KBr"))) * "119.0 g"/(1color(red)(cancel(color(black)("mole KBr")))) = "549.78 g"`

Finally, a solution's percent concentration by mass is definedas the mass of solute divided by the total mass of the solution, and multiplied by `100`

`color(blue)("% w/w" = "mass of solute"/"mass of solution" xx 100)`

Since this `"1.00-L"` sample has a mass of `"1370 g"`, the solution's percent concentration by mass will be

`"% w/w" = (549.78 color(red)(cancel(color(black)("g"))))/(1370color(red)(cancel(color(black)("g")))) xx 100 = color(green)("40.1%")`

The answer is rounded to three sig figs.