Molarity of 56.6% Ammonium Hydroxide (or 28% aqueous solution of ammonia) [NH4OH (NH3 + H2O)]

Molarity of 56.6% Ammonium Hydroxide (or 28% aqueous solution of ammonia) [NH4OH (NH3 + H2O)]

Overview:

  • Ammonium Hydroxide (NH4OH) is a clear colourless liquid. 56.6% (w/w) concentrated Ammonium Hydroxide can be obtained from different suppliers.
  • When ammonia dissolves in water, it forms Ammonium Hydroxide. 28% aqueous solution of Ammonia (NH3) is equal to approximately 56.6% Ammonium Hydroxide.
  • 56.6% (w/w) Ammonium Hydroxide means that 100 grams of Ammonium Hydroxide contains 56.6 grams of NH4OH.
  • The density of 56.6% (w/w) Ammonium Hydroxide is 0.9 g/ml at 25°C that means that weight of the 1 ml of Ammonium Hydroxide is 0.9 gram at 25°C.
  • Molarity refers to number of moles of the solute present in 1 litre of solution.
  • In simple words, 1 mole is equal to atomic weight of the substance. For example, 1 mole of NH4OH is equal to 35.05 grams of NH4OH (molecular weight = 35.05).

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Calculation procedure:

Known values:
Density of Ammonium Hydroxide :                        0.9 g/ml
Molecular weight of NH4OH:                                  35.05 g/mole

Concentration of Ammonium Hydroxide:            56.6% (% by mass, wt/wt)

Step 1: Calculate the volume of 100 grams of Ammonium Hydroxide.

Formula:
Density =  weight / volume    or
Volume =  weight / density    or

Volume of 100 gram of Ammonium Hydroxide : 100/0.9 =  111.11 ml

Note: 56.6% (w/w) Ammonium Hydroxide means that 100 grams of Ammonium Hydroxide contain 56.6 grams of NH4OH.

The volume of 100 grams of Ammonium Hydroxide is 111.11 ml. That means 56.6 grams of NH4OH is present in 111.11 ml of Ammonium Hydroxide.

Step 2: Calculate how many grams of NH4OH is present in 1000 ml of Ammonium Hydroxide.

111.11 ml of Ammonium Hydroxide contain       = 56.6 grams of NH4OH
1 ml of Ammonium Hydroxide will contain           = 56.6/111.11 grams of NH4OH

1000 ml of Ammonium Hydroxide will contain     = 1000 x 56.6/111.11 = 509.405 grams of NH4OH

1000 ml of Ammonium Hydroxide will contain 509.405 grams of NH4OH

Step 3: Calculate the number of moles of NH4OH present in 509.405 grams of NH4OH.

35.05 grams of NH4OH is equal to 1 mole.
1 gram of NH4OH will be equal to 1/35.05 moles.

509.405 grams will be equal to = 509.405 x 1/35.05 = 14.534 moles

Therefore, we can say that 1 litre of Ammonium Hydroxide contains 14.534 moles or in other words molarity of 56.6% (w/w) Ammonium Hydroxide is equal to 14.534 M.