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Molality (m), Equation 5.7

>> Parts of this equation/concept include:

A -	Determining Molality from Solute and Solvent
B -	Determing Solute from Molality and Solvent
C -	Converting between Molarity and Molality

molality = moles of solute/kilograms of solvent

Questions for problems involving molality will be phrased slightly differently, since you will need solvent instead of solution. It will normally give mass of solvent, but it might give density. Be careful to notice whether it is density of solvent or density of solution, since density does not change the type of substance (such as concentration units); it only changes mass to volume. Also recall that the density of water, the most common solvent, is 1.00 g/mL.

Also note: Unlike volumes, masses are additive. The following equation might be useful:

g solution = g solvent + g solute

>> Example 1

What is the molality of a solution made by dissolving 0.1356 g MgSO₄ in 200.0 mL of water?

Solution:

molality = mol solute/kg sol
solute = MgSO₄
solvent = water

moles of solute

=

0.1356 g

1 mol 120.367 g

=

1.12710–3 mol

kg of solvent

=

200.0 mL

1.000 g 1 mL

1 kg 1000 g

=

0.2000 kg

molality

=

1.12710–3 mol 0.2000 kg

=

5.63310–3 m

If you use a density of water values with fewer than four significant figures, the number of significant figures in the overall answer will be the same as that in the density value used.

>> Example 2

What is the molality of a solution of 6.44 g of Cd(NO₃)₂ in 375.0 g of water?

Solution:

molality = mol solute/kg solvent
solute = Cd(NO₃)₂
solvent = water

moles of solute

=

6.44 g

1 mol 236.42 g

=

0.0272 mol Cd(NO3)2

kg of solvent

=

375.0 g

1 kg 1000 g

=

0.3750 kg

molality

=

0.0272 mol 0.3750 kg

=

0.0726 m

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>> Example 3

How many grams of Sr(ClO₄)₂ are required to make a 0.30 m aqueous solution using 600 g of water?

Solution:

molality = mol solute/kg solvent
solute = Sr(ClO₄)₂
solvent = water

kg solvent

=

600 g water

1 kg 1000 g

=

0.6000 kg

mol solute	=	molality		kg solvent
	=	0.30 m		0.6000 kg
	=	0.18 mol Sr(ClO4)2

g solute

=

0.18 mol Sr(ClO4)2

286.52 g 1 mol

=

52 g

or (equally correct!)

600 g water

1 kg 1000 g

0.30 mol Sr(ClO4)2 1 kg water

286.52 g 1 mol Sr(ClO4)2

=

52 g

>> Example 4

How many grams of BiCl₃ are needed to make 500.0 g of a 0.10-m aqueous solution?

Solution:

molality = mol solute/kg solvent
solute = BiCl₃
solvent = water

Note that this is g of solution, not of solvent! However, g solvent = g solution – g solute.

Note that g solvent/1000 = kg solvent or (g solution – g solute)/1000 = kg solvent.

Note that mole of solute = (g solute)/(molar mass). The molar mass of BiCl₃ is 315.34 g/mol.

If we let x = g solute; then the molality equation can be rewritten as

0.10 m

=

x/315.34 (500.0 – x)/1000

0.0500 – (110–4)x

=

x 315.34

15.767 – 0.0315x = x

15.767 = 1.0315x

15.28 = x

or 15 g of BiCl₃ (rounded to the appropriate significant figures).

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In problems of conversion between molality and molarity you will need to be given the density of the solution. Since this varies with both the identity and the concentration of the solution, it must be given as part of the problem. You will need it to convert from mass in molality to volume in molarity. Note also that not only do the units change, but the material also changes from solution (in molarity) to solvent (in molality).

The concentration of a solution does not depend on how much solution is present. You will probably find it convenient to choose a particular amount of solution. The most convenient thing is to choose 1 L of solution if you are starting with molarity and 1 kg of solvent if you are starting with molality.

>> Example 5

What is the molality of a 0.123 M HCl(aq) solution? The density of the solution is 1.030 g/mL.

Solution:

molality = mol HCl/kg water
molarity = mol HCl/L solution

Since we have a value of molarity given, let's assume that we are working with exactly 1 L of solution. Using the molarity equation M = mol/L or ML = mol, there is (0.123 M)(1 L) = 0.123 mol HCl.

0.123 mol HCl

36.45 g 1 mol

=

0.0448 g HCl

Using the density

1 L

1000 mL 1 L

1.030 g 1 mL

=

1030 g solution

g solvent = g solution – g solute = 1003 g – 0.0448 g = 1030 g solvent

1030 g solvent

1 kg 1 g

=

1.030 kg

Using the molality equation

m

=

mol HCl kg solvent

=

0.123 mol 1.030 kg

=

0.119 m

>> Example 6

What is the molarity of a 3.21 m KOH(aq) solution? (density of solution = 1.163 g/mL)

Solution:

M = mol KOH/L solution
m = mol KOH/kg water

Since we have a value for molality, let's assume exactly 1 kg (1000 g) of water.

mkg = mol KOH = (3.21 m)(1 kg) = 3.21 mol KOH

3.21 mol KOH

56.10 g 1 mol

=

180.1 g

g solution = g solvent + g solute = 1000 g + 180.1 g = 1180.1 g

1180.1 g solution

1 mL 1.163 g

1 L 1000 mL

=

1.015 L

M

=

mol KOH L solution

=

3.21 mol KOH 1.015 L

=

3.16 M

Notice that although the values for molality and molarity are similar, the value for molality will be slightly less. The difference normally becomes larger at higher concentrations.

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