Help calculating molality?

Calculate the molality of HCL

Formula weight (amu) 36.465
Density 1.19 g/mL
Weight % 37.2
Molarity 12.1

I already did it but I just want to make sure it's correct. Thanks!

Answer:
We consider 1.00 L of this solution.
Mass of 1.00 L = 1190 g
1190 g/L x 37.2 / 100 = 442.7 g/L of HCl
442.7 g /L / 36.465 = 12.1 M ( this is the method to find molarity )
In 1.0 L of this solution there are 12.1 moles
12.1 mol x 36.465 g/mol = 441.2 g of HCl
1190 - 441.2 = 748.8 g of solvent = 0.7488 Kg
m = moles solute / Kg solvent = 12.1 mol / 0.7488 = 16.1 m
IT' S CORRECT

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