How many grams of solute are present in 605 mL of 0.320 M KBr?

Answer 1

The solute is KBr or potassium bromide. The molecular weight is 118.99 gr/mol. Using the molarity given (0.32 mol/L) multiply them together i.e. 118.99 g 0.32 mol 38.07 KBr g

----------- X ------------- = ------------------- (the mol unit labels cancel)

mol L L

Your only using 650 ml of solution so: 1L 38.07 g .03807 g KBr

------------- X -------------- = ----------- X 650 ml = 24.74 gr KBr

1,000 ml 1 L ml

all the units cancel and your left with grams of KBr, the solute