As in the HCl problem, start with the definition.
m = mols/kg solvent
mass of 1 L = 1000 mL x 1.38 g/mL = 1380 g.
Not all of that is HBr. HBr present is 40% so 1380 x 0.40 = 552g.
mols in 552g HBr is mols = grams/molar mass = 552/about 81 (but you need to do it more accurately).
mols HBr = estd 6.81.
How much solvent is there? You have 1380g solution of which 552g is HBr; therefore, g solvent 1380-552 = 828g.
Convert 828g to kg, then m = mols/kg solvent = ?m
Commercial hydrobromic acid can be purchased as 40.0 % by mass 9f HBr. The density of this acid is 1.38 g m/L. Calculate the molality of this commercial acid..please show me the solution..thanks
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