The concentration of kbr in a solution prepared by dissolving 2.10 g of kbr in 897 g of water is molal.

Molality is defined as the number of moles of solute in 1 kg of solvent.
In this case the solvent is water, therefore we need to find the number of moles of KBr in 1 kg of water.
Number of moles of KBr - 2.10 g / 119 g/mol  = 0.0176 mol
mass of water - 897 g
number of moles of KBr in 897 g - 0.0176 mol
Therefore number of KBr moles in 1000 g - 0.0176 mol / 897 g x 1000 g = 0.0196 mol/kg
therefore molality of KBr is 0.0196 mol/kg

Ithinks its false. tell me if im wrong

Ithink this is what you meant. the temperature at which a liquid begins to boil is called the boiling point. the boiling point of water is 212°f or 100°c.  and at that point will start to turn into steam (gas)