### MolalityTen Examples

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.

 moles of solute Molality = ––––––––––––––––– 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:

 1.00 mol Molality = –––––––– 1.00 kg

Notice that both the units of mol and kg remain. Neither cancels.

A symbol 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. (A lower-case m is also used for meter, but the context should be clear that m means molality.) 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 a symbol that stands for mol/kg. It is not the actual unit.

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

 2.00 mol Molality = –––––––– 1.00 kg

Notice that no mention of a specific substance is mentioned at all. The molarity would be the same no matter what the substance. 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 0.750 mol is dissolved in 2.50 L of solvent?

 0.750 mol Molality = ––––––––– 2.50 kg

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 molar mass of NaCl.

Step One: 58.44 g / 58.44 gr/mol = 1.00 mol.

Step Two: 1.00 mol / 2.00 kg = 0.500 mol/kg (or 0.500 m).

Sometimes, a book will write out the word "molal," as in 0.500-molal.

Example #5: Calculate the molality of 25.0 grams of KBr dissolved in 750.0 mL pure water.

 25.0 g ––––––––– = 0.210 mol 119.0 g/mol

 0.210 mol ––––––––– = 0.280 m 0.750 kg

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

 80.0 g ––––––––– = 0.444 mol 180. g/mol

 0.444 mol ––––––––– = 0.444 m 1.00 kg

Example #7: Calcuate the molality when 75.0 grams of MgCl2 is dissolved in 500.0 g of solvent.

Rather than a two-step approach, some teachers will want you to combined the steps into one diagram. This technique is called dimensional analysis.

 1 mol 1 75.0 g x ––––––– x ––––––– = 1.58 m 95.2 g 0.500 kg

Example #8: 100.0 grams of sucrose (C12H22O11, mol. wt. = 342.3 g/mol) is dissolved in 1.50 L of water. What is the molality?

 1 mol 1 100.0 g x ––––––– x ––––––– = 0.195 m 342.3 g 1.500 kg

Example #9: 49.8 grams of KI is dissolved in 1.00 kg of solvent. What is the molality?

Two step:

 49.8 g ––––––––– = 0.300 mol 166.0 g/mol

 0.300 mol ––––––––– = 0.300 m 1.00 kg

Dimensional analysis:

 1 mol 1 49.0 g x ––––––– x ––––––– = 0.300 m 166.0 g 1.00 kg

Example #10: You have 849.0 grams of water and you wish to make a 5.2 m (molality) solution of mercury(II) oxide, HgO. How many grams of the solute would you have to add to the water that you have?

 5.2 mol 0.849 kg 216.589 g ––––––– x ––––––– x ––––––– = 956 g 1.00 kg 1 1 mol

HgO is rather insoluble in water, so the above solution could not be made. of course, the point of the question is not to make the solution, it is to learn how to do molality calculations.

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. The words differ by only one letter and sometimes that small difference is overlooked.

Bonus Example: The density of a solution of 3.69 g KCl in 21.70 g H2O is 1.11 g/mL. Calculate the molality of KCl in the solution.

Solution:

1) Determine moles of KCl:

3.69 g / 74.551 g/mol = 0.049496 mol

2) Determine the molality:

0.049496 mol / 0.02170 kg = 2.28 m

3) Note that the density is not needed. It's a red herring.