A 2.500 g portion of sodium hydroxide (NaOH) is placed in a 250.00 mL volumetric flask and diluted to the water mark. It is later found that the added mass of water was 497.2 g. What is the final concentration of NaOH in units of molarity and molality?
To calculate the molarity (M) of NaOH, we first need to calculate the number of moles of NaOH present in the solution:
moles of NaOH = mass / molar mass = 2.500 g / 40.00 g/mol = 0.06250 mol
Next, we need to calculate the volume of the solution in liters:
volume = 250.00 mL / 1000 mL/L = 0.2500 L
Now we can use the formula for molarity:
M = moles of solute / liters of solution
M = 0.06250 mol / 0.2500 L = 0.2500 M
Therefore, the final concentration of NaOH is 0.2500 M.
To calculate the molality (m) of NaOH, we need to consider the mass of the solvent (i.e. water) instead of the volume of the solution. The mass of the solvent can be calculated by subtracting the mass of the solute (NaOH) from the total mass of the solution:
mass of solvent = total mass of solution - mass of solute
mass of solvent = 497.2 g - 2.500 g = 494.7 g
We can now use the formula for molality:
m = moles of solute / mass of solvent (in kg)
m = 0.06250 mol / 0.4947 kg = 0.1263 m
Therefore, the final concentration of NaOH is 0.1263 m.