If you have 670.0 g of water and wish to make a 2.13 m solution of kbr, how many grams of the solute would you have to add to the water that you have?

I believe that for this item, the unit m is for the molality which is equal to mole of solute to kg of solvent. For water that is 670 g, the equivalent in kg is equal to 0.670 kg. Then using the equation,
                          molality = moles solute / kg solvent
and substituting the known values,
                        2.13 = moles solute / 0.670 kg
The amount of solute should be 1.4271 moles. Then, we multiply this value with the molar mass of KBr which is equal to 119.002 g/mol. The answer would be 169.83 g.

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