The density of 28% ammonium hydroxide is approximately 0.9 g/mL. This means that 100 mL of 28% NH4OH weighs 90 g.
The molar mass of NH4OH is 35 g/mol.
Therefore, the number of moles of NH4OH in 90 g of 28% NH4OH is:
Moles of NH4OH = (90 g)/(35 g/mol) = 2.57 moles
The volume of 90 g of 28% NH4OH is:
Volume = (90 g)/(0.9 g/mL) = 100 mL = 0.1 L
Now we can calculate the normality of 28% ammonium hydroxide:
Normality = (Moles of NH4OH)/(Volume in liters)
Normality = (2.57 moles) / (0.1 L) = 25.7 N
Therefore, the normality of 28% ammonium hydroxide is approximately 25.7 N.
