If 80.0 g of NaOH is dissolved in 1 kg of water, what is the molality of the solution?
The molality of a solution is calculated like this:
molality = moles of solute / mass of solvent in kg
The solute of our solution is NaOH and the solvent is water. We already know the mass of solvent in kg.
Mass of solvent in kg = 1 kg
To find the molality we still have to find the number of moles of solute. To do that we will have to find the molar mass of NaOH, and then use it to find the number of moles of NaOH that we have in 80.0 g of it.
atomic mass of Na = 22.99 amu
atomic mass of O = 16.00 amu
atomic mass of H = 1.01 amu
molar mass of NaOH = 1 * 22.99 + 1 * 16.00 + 1 * 1.01
molar mass of NaOH = 40.00 g/mol
number of moles of NaOH = 80.0 g/(40-0 g/mol)
number of moles of NaOH = 2.00 moles
Now that we know the number of moles of NaOH and the mass of solvent in kg we can calculate the molality of the solution:
molality = moles of solute / mass of solvent in kg
molality = 2.00 moles/1 kg
molality = 2.00 m
Answer: the molality of the solution is 2.00 m