How do you calculate the molarity of a solution prepared by dissolving 15.6 g of solid KBr in enough water to make 1.25 L of solution?

For starters, convert the mass of potassium bromide to moles by using the molar mass of the compound.

`15.6 color(red)(cancel(color(black)("g"))) * "1 mole KBr"/(119.002color(red)(cancel(color(black)("g")))) = "0.1311 moles KBr"`

Now, in order to find the molarity of the solution, you need to figure out how many moles of solute are present for every `"1 L"` of solution.

You already know that `"1.25 L"` of solution contain `0.1311` moles of potassium bromide, the solute, so all you have to do now is to use this known composition as a conversion factor to go from `"1 L"` of solution to number of moles of solute.

`1 color(red)(cancel(color(black)("L solution"))) * "0.1311 moles"/(1.25color(red)(cancel(color(black)("L solution")))) = "0.10488 moles KBr"`

Since this represents the number of moles of solute present in `"1 L"` of solution, you can say that the molarity of the solution is equal to

`color(darkgreen)(ul(color(black)("molarity = 0.105 mol L"^(-1))))`

The answer must be rounded to three sig figs, the number of sig figs you have for your values.