The molarity of NaBr = 0.05036M
0.0460 m NaBr has 0.0460 moles NaBr with 1000g of solvent
therefore the mass of NaBr= 0.0460 moles x 102.89 g/mol =4.73grams
total mass is therefore= 4.733 g +1000= 1004.73 grams
step 2:
use the density to determine the volume
vol=mass/density
=1004.73 g/1.10 g/mol=913.39 ml in liters= 913.39/1000=0.91339 L
now the molarity = moles/vol in liters
=0.0460 m/0.91339=0.05036 M
