Question

Question #5c6f3

Answer 1

You need `"105 mL"` of your stock potassium bromide solution to get that much potassium bromide.

Molarity is defined as moles of solute, in your case potassium bromide, divided by liters of solution.

The first thing you need to do is figure out how many moles of potassium bromide you have in 2.50 g - use the compound's molar mass to determine this

`2.50cancel("g") * "1 mole KBr"/(119.002cancel("g")) = "0.021 moles KBr"`

Next, try to figure out exactly how much volume of your 0.200 M solution would have that many moles. Since concentration is moles per volume, you can solve for the volume by

`C = n/V => V = n/C`

`V = (0.021cancel("moles"))/(0.200cancel("mol")/"L") = "0.105 L"`

Expressed in mililiters, the answer will be

`0.105cancel("L") * ("1000 mL")/(cancel("1 L")) = color(green)("105 mL")`

So, as a conclusion, when dealing with molarities, always try to have moles and liters. If a mass of solute is given, like it was in this case, immediately convert it to moles by using its molar mass

In your case, 105 mL of a `KBr` solution that has 0.200 moles per liter will contain the number of moles of `KBr` you have in 2.50 g.