One can take 15 M NaOH as equivalent to 15 mol/L NaOH. We know that density equals mass/Volume, in SI units of kg/m3 or kg/L for aqueous, liquids, or solids. Therefore, 15 mol/L of NaOH can be stoichiometrically converted to grams/L and then to kg/L which would give us the SI units of density.
Looking at the periodic table, we can determine the overall molecular weight or g/mol of NaOH by summing the g/mol of each atomic species, Na is ~23 g/mol, O is ~16.00g/mol, and H is ~ 1.0 g/mol. The total molecular weight or amu of NaOH is thus ~40g/mol.
Stoichiometry Table:
15 mol/L NaOH x (40.0 g/mol) --> cancels the moles and leaves us with 600 g/L NaOH
which may subsequently be converted into kg/L, the SI unit of density:
600 g/L x (1kg/1000g) = 0.6 kg/L. Thus, the density is 0.6 kg/L or 0.6kg/m3.
Let me know if there are any further concerns or questions.
Benjamin C.
J.R.S. is right. This is a nasty question if one is going to answer it correctly. In dilute solutions one could assume that the density would be that of water PLUS the mass of the NaOH added, but in seriously concentrated solutions like this one, the amount of actual H2O per liter starts going down. Probably, whomever is asking the question is looking for the mass of NaOH PLUS the mass of water at Standard Temperature and Pressure (STP), but this wouldn't be actually correct. But assuming that is what they want, NaOH has a molar mass of 22.990+15.999+1.0080 = 39.997 g/l, and a 15 M solution of that would have 599.955 g/liter - adjusted to 600.00 g/l to keep significant digits correct, which gives us 0.60000 kg/l. This doesn't actually change the answer that Sean R. gave, except for managing significant digits.07/24/23
J.R. S.
07/24/23