Question #5c6f3

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.

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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

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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.