What is the molality of a solution containing 11.1 g Na2SO4 dissolved in 2269.2 g of water?
The molality formula is mol solute/kg solvent. In this case, Na2SO4 is the solute and water is the solvent.
So basically you have to find the moles of Na2SO4 by dividing 11.1 g by molar mass of the solute, which is about 142 g/mol (11.1g/ 142 g/mol)
Then you convert grams of water into kilograms. You can multiply by the conversion factor, which is essentially dividing the amount of 1000. You get 2.2692 kg
The molality would be
(11.1g/142 g/mol)/2.4448 kg. The answer is expressed as a value of "m"
answer should be 0.0344