Densities of solutions of sodium hydroxide and water at 20 °C are given by the equation: p 1.003 + 1.06Wao where p is the density of the solution in g/cm (grams of solution per cm3 of solution) and wNaOH is the weight fraction of sodium hydroxide. The equation is valid for sodium hydroxide weight fractions less than or equal to 0.5. 100 liters of a solution that has a sodium hydroxide weight fraction of 0.5 is to be diluted with pure water until the weight fraction of NaOH is equal to 0.2. What volume of water must be added and what will be the volume of the final mixture? Express both volumes in liters. NaOH

50.08%

Explanation:

Hydroxide sodium is a basis, so it's molecule is NaOH (an Arrhenius basis is a cation followed by OH, in this case, sodium is the cation).

The question wants to know the mass percent, so we can choose any calculus basis because the fraction will be the same. So, for convenience, choosing the total number of moles as 1 mol.

The molar fraction of NaOH is 0.3110, so it's the number of moles on the basis is 0.3110, and the rest of the solution is water, which is 0.6890 moles.

The molar mass of NaOH is 40 g/mol (Na = 23 g/mol, O = 16 g/mol, H = 1 g/mol) and the molar mass of water (H2O) is 18 g/mol.

So, calculating the mass by the equation:

m = nxMM

Where m is the mass, n is the number of moles and MM is the molar mass.

NaOH: m = 0.3110x40 = 12.4400 g

H2O: m = 0.6890x18 = 12.4020 g

The total mass of the solution is 24.8420 g, and the mass percent (x) of NaOH is:

x = 50.08%