The molarity of 98% (w/w) Ethanolamine (ETA) is 16.237 M.

A 98% (w/w) Ethanolamine is a clear colorless solution. It can be purchased from many suppliers.

The “%” refers to solution concentration in percentage and “(w/w)” refers to solute and solution amount given in grams (i.e., percentage by weight). This means 98% (w/w) Ethanolamine solution contains 98 g of Ethanolamine per 100 g of solution.

The density of 98% (w/w) ethanolamine solution is 1.012 g/ml at 25°C, meaning the weight of the 1 ml of the ethanolamine solution is 1.012 g at 25°C.

Molarity refers to the number of moles of the solute present in 1 liter of solution. To calculate the Molarity of 98% (w/w) Ethanolamine, you need to know how many moles of Ethanolamine are in 1 liter of solution.

In simple words, 1 mole is equal to the molecular weight of the substance. For example, 1 mole of Ethanolamine is equal to 61.08 g of Ethanolamine (Molecular weight: 61.08). To calculate the molarity, one must first calculate how much Ethanolamine is present in 1 L of 98% solution. Once we know the amount of Ethanolamine in 1 L solution, we can calculate the molarity of the solution by dividing the Ethanolamine amount by the molecular weight.

**Calculation procedure: **Molarity of 98% (w/w) Ethanolamine (ETA)

Known values | |

Molar mass of Ethanolamine | 61.08 g/mole |

Concentration of Ethanolamine | 25% (% by mass, w/w) |

Density of 98% Ethanolamine | 1.012 g/ml |

**Step 1: Calculate the volume of 100 grams of Ethanolamine**

Formula:

Density = Weight/Volume

OR

Volume = Weight/Density

The volume of 100 g of Ethanolamine solution: 100/1.012 = 98.8142 ml

Note: 98% (w/w) Ethanolamine solution means 100 g of solution contains 98 g of Ethanolamine.

The volume of 100 g of Ethanolamine solution is 98.8142 ml, meaning 98 g of Ethanolamine is present in 98.8142 ml of Ethanolamine solution.

**Step 2: Calculate how many grams of Ethanolamine is present in 1000 ml of Ethanolamine solution.**

98.8142 ml of Ethanolamine solution contains = 98 grams of Ethanolamine

1 ml of Ethanolamine solution will contain = 98/98.8142 grams of Ethanolamine

1000 ml of Ethanolamine solution will contain = (1000 x 98)/98.8142 = 991.76 grams of Ethanolamine

1000 ml of Ethanolamine solution will contain 991.76 grams of Ethanolamine.

**Step 3: Calculate the number of moles of Ethanolamine in 991.76 grams of Ethanolamine solution.**

61.08 g of Ethanolamine is equal to 1 mole.

1 g of Ethanolamine will be equal to 1/61.08 moles.

991.76 grams of Ethanolamine will be equal to = (991.76 x 1)/61.08 = 16.237 moles

Therefore, we can say that 1 liter of Ethanolamine solution contains 16.237 moles of Ethanolamine or in other words molarity of 98% (w/w) Ethanolamine solution is equal to 16.237 M.

## Molarity of 98% (w/w) Ethanolamine

**Calculator – Calculate the molarity of concentrated Ethanolamine (ETA)**

**Ethanolamine (ETA) Molecular weight:** 61.08 g/mole

**Concentration of Ethanolamine:**% (wt/wt)

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**Density of Ethanolamine solution:** g/ml

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**Molarity of Ethanolamine solution:** 16.237 M