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As is clear from its name, molality involves moles. Boy, does it!

The molality of a solution is calculated by taking the moles of solute and dividing by the kilograms of solvent.

This is probably easiest to explain with examples.

Example #1 - Suppose we had 1.00 mole of sucrose (it's about 342.3 grams) and proceeded to mix it into exactly 1.00 liter water. It would dissolve and make sugar water. We keep adding water, dissolving and stirring until all the solid was gone. We then made sure everything was well-mixed.

What would be the molality of this solution? Notice that my one liter of water weighs 1000 grams (density of water = 1.00 g / mL and 1000 mL of water in a liter). 1000 g is 1.00 kg, so:

The answer is 1.00 mol/kg. Notice that both the units of mol and kg remain. Neither cancels.

A replacement for mol/kg is often used. It is a lower-case m and is often in italics, m. Some textbooks also put in a dash, like this: 1.00-m. However, if you write 1.00 m for the answer, without the italics, then that usually is correct because the context calls for a molality. Having said that, however, be aware that often m is used for mass, so be careful. Maybe including the dash would be wise if there might be a potential misunderstanding

When you say it out loud, say this: "one point oh oh molal." You don't have to say the dash.

And never forget this: replace the m with mol/kg when you do calculations. The m is just shorthand for mol/kg.

Example #2 - Suppose you had 2.00 moles of solute dissolved into 1.00 L of solvent. What's the molality?

The answer is 2.00 m.

Notice that no mention of a specific substance is mentioned at all. The molarity would be the same. It doesn't matter if it is sucrose, sodium chloride or any other substance. One mole of anything contains 6.022 x 1023 units.

Example #3 - What is the molality when 0.75 mol is dissolved in 2.50 L of solvent?

The answer is 0.300 m.

Now, let's change from using moles to grams. This is much more common. After all, chemists use balances to weigh things and balances give grams, NOT moles.

Example #4 - Suppose you had 58.44 grams of NaCl and you dissolved it in exactly 2.00 kg of pure water (the solvent). What would be the molality of the solution?

The solution to this problem involves two steps.

Step One: convert grams to moles.

Step Two: divide moles by kg of solvent to get molality.

In the above problem, 58.44 grams/mol is the molecular weight of NaCl. (For you technical types, I know it actually is a formula weight, but I'm glossing over the difference for the time being. Remember, this is a high school tutorial.)

Dividing 58.44 grams by 58.44 grams/mol gives 1.00 mol.

Then, dividing 1.00 mol by 2.00 kg of solvent gives 0.500 mol/kg (or 0.500 m). Sometimes, a book will write out the word "molal," as in 0.500-molal.

Do examples #5 and #6:

5) Calculate the molality of 25.0 grams of KBr dissolved in 750.0 mL pure water.

6) 80.0 grams of glucose (C6H12O6, mol. wt = 180. g/mol) is dissolved in1.00 kg of water. Calculate the molality.

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Practice Problems

1) Calcuate the molality when 75.0 grams of MgCl2 is dissolved in 500.0 g of solvent.

2) 100.0 grams of sucrose (C12H22O11, mol. wt. = 342.3 g/mol) is dissolved in 1.50 L of water. What is the molality?

3) 49.8 grams of KI is dissolved in 1.00 kg of solvent. What is the molality?

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In the molarity tutorial the phrase "of solution" kept showing up. The molarity definition is based on the volume of the solution. This makes molarity a temperature-dependent definition. However, the molality definition does not have a volume in it and so is independent of any temperature changes. This will make molality a very useful concentration unit in the area of colligative properties.

Lastly, it is very common for students to confuse the two definitions of molarity and molality. They differ by only one letter and sometimes that small difference is overlooked.

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