Molality Formula: Definition, SI Unit, Derivation and Solved Examples

The molality formula is $m = \dfrac{n}{W}$, where $m$ is the molality of the solution, $n$ is the number of moles of solute, and $W$ is the mass of the solvent in kilograms. Molality is expressed in $\mathrm{mol\ kg^{-1}}$ or molal (m). It tells us the amount of solute present per kilogram of solvent, and unlike molarity, it does not change with temperature. That is why the molality formula is used in colligative property numericals, such as boiling point elevation and freezing point depression, in Class 11, Class 12, JEE and NEET.

Molality Formula at a Glance

Choose the form of the molality formula according to the data that the question gives.

What Is Molality?

Molality is the number of moles of solute dissolved in one kilogram of solvent.

$m = \dfrac{\text{Moles of solute}}{\text{Mass of solvent in kilograms}}$

Therefore,

$m = \dfrac{n}{W}$

Where:

A solution with a molality of $1\ \mathrm{mol\ kg^{-1}}$ is called a 1 molal solution and is written as $1\ \mathrm{m}$.

Important Points About Molality

• Molality depends on the mass of the solvent, not the mass or volume of the solution.
• The mass used must be the mass of the solvent alone, not the total solution.
• If the mass of solvent is given in grams, it must be converted to kilograms before applying $m = \dfrac{n}{W}$.
• Molality is temperature-independent because mass does not change with temperature, unlike volume.
• Molality is commonly used in colligative property calculations, cryoscopy, and ebullioscopy.

For a deeper grounding in moles before working through these formulas, see the Mole Concept notes for Class 11, IIT JEE & NEET.

What Is the Formula of Molality?

The formula of molality is:

$m = \dfrac{n}{W}$

Here:

• $m$ = molality of solution in $\mathrm{mol\ kg^{-1}}$
• $n$ = number of moles of solute
• $W$ = mass of solvent in kilograms

If the number of moles is not given directly, it can be calculated using:

$n = \dfrac{w}{M_{\mathrm{molar}}}$

where:

• $w$ = mass of solute in grams
• $M_{\mathrm{molar}}$ = molar mass of solute in $\mathrm{g\ mol^{-1}}$

Substituting this value in the molality formula:

$m = \dfrac{w}{M_{\mathrm{molar}} \times W}$

where $W$ is in kilograms.

How to Calculate Molality: Step by Step

1. Write the mass of the solute in grams and the mass of the solvent in grams.
2. Find the moles of solute: $n = \dfrac{w}{M_{\mathrm{molar}}}$, using the molar mass of the solute.
3. Convert the mass of the solvent to kilograms: divide the mass in grams by 1000.
4. Apply the molality formula: $m = \dfrac{n}{W}$, and write the answer in $\mathrm{mol\ kg^{-1}}$ (molal).

Molality Formula When Mass of Solute Is Given

When the mass of the solute is given in grams and the mass of the solvent is given in grams as well, the commonly used molality formula is:

$m = \dfrac{w}{M_{\mathrm{molar}}} \times \dfrac{1000}{W_{\mathrm{g}}}$

Where:

• $w$ = mass of solute in grams
• $M_{\mathrm{molar}}$ = molar mass of solute in $\mathrm{g\ mol^{-1}}$
• $W_{\mathrm{g}}$ = mass of solvent in grams

Example: Molality Using Mass of Solute

Suppose $10\ \mathrm{g}$ of $\ce{NaOH}$ is dissolved in $500\ \mathrm{g}$ of water.

Molar mass of $\ce{NaOH}$:

$M_{\mathrm{molar}} = 40\ \mathrm{g\ mol^{-1}}$

Therefore,

$m = \dfrac{10}{40} \times \dfrac{1000}{500}$

$m = 0.25 \times 2 = 0.5\ \mathrm{m}$

Hence, the molality of the $\ce{NaOH}$ solution is:

$\boxed{0.5\ \mathrm{m}}$

Molality Formula When the Mass of Solution Is Given

In many numericals, the question gives the mass of the solution (for example “a 20% by mass solution”) and not the mass of the solvent. Find the mass of the solvent first:

$\text{Mass of solvent} = \text{Mass of solution} - \text{Mass of solute}$

Then use the molality formula with this mass of solvent. This is the most common place where students lose marks, because they divide by the mass of the solution.

Example: Find the molality of a 20% by mass aqueous solution of glucose ($M_{\mathrm{molar}} = 180\ \mathrm{g\ mol^{-1}}$).

In $100\ \mathrm{g}$ of solution there is $20\ \mathrm{g}$ of glucose, so the mass of water is $100 - 20 = 80\ \mathrm{g} = 0.08\ \mathrm{kg}$.

$n = \dfrac{20}{180} = 0.111\ \mathrm{mol}$

$m = \dfrac{0.111}{0.08} \approx 1.39\ \mathrm{mol\ kg^{-1}}$

$\boxed{m \approx 1.39\ \mathrm{m}}$

The mass percent form of the molality formula gives the same answer: $m = \dfrac{20 \times 1000}{180 \times (100 - 20)} = 1.39\ \mathrm{m}$.

SI Unit and Symbol of Molality

The unit of molality is $\mathrm{mol\ kg^{-1}}$.

Note: the unit $\mathrm{mol\ kg^{-1}}$ is made only of SI base units (mole and kilogram), while the common unit of molarity, $\mathrm{mol\ L^{-1}}$, is not. Molality is preferred in precise work because it does not change with temperature.

Derivation of Molality Formula

The definition of molality is:

$m = \dfrac{\text{Number of moles of solute}}{\text{Mass of solvent in kilograms}}$

The number of moles of solute is:

$n = \dfrac{w}{M_{\mathrm{molar}}}$

Substituting this into the molality equation:

$m = \dfrac{w}{M_{\mathrm{molar}} \times W}$

where $W$ is in kilograms.

If the mass of solvent is given in grams:

$W(\mathrm{kg}) = \dfrac{W_{\mathrm{g}}}{1000}$

Therefore,

$m = \dfrac{w}{M_{\mathrm{molar}} \times \left(\dfrac{W_{\mathrm{g}}}{1000}\right)}$

Hence,

$\boxed{m = \dfrac{w}{M_{\mathrm{molar}}} \times \dfrac{1000}{W_{\mathrm{g}}}}$

This is the most commonly used expanded molality formula in numerical problems.

Molality vs Molarity vs Mole Fraction

Molarity, molality, and mole fraction are different ways of expressing the concentration or composition of a solution.

Molarity numericals follow the same solutions framework, so see the complete Molarity Formula guide for the volume-based approach and the dilution equation.

Key Difference Between Molality and Molarity

Since mass does not change appreciably with temperature, molality stays constant when temperature changes. Volume, on the other hand, expands on heating, so molarity decreases as temperature rises. This is exactly why the colligative property laws, $\Delta T_b = K_b \times m$ and $\Delta T_f = K_f \times m$, are defined using molality and not molarity.

Relation Between Molarity and Molality

If the molarity ($M$), the density of the solution ($d$, in $\mathrm{g\ mL^{-1}}$), and the molar mass of the solute ($M_{\mathrm{molar}}$) are known, molality can be calculated as:

$\boxed{m = \dfrac{1000M}{1000d - M \times M_{\mathrm{molar}}}}$

In the other direction, molarity from molality is:

$\boxed{M = \dfrac{1000\, m\, d}{1000 + m \times M_{\mathrm{molar}}}}$

These relations are frequently asked in JEE Main and JEE Advanced, especially in problems where density is given instead of the mass of solvent directly. For dilute aqueous solutions, molality and molarity are almost equal, because the density of the solution is close to $1\ \mathrm{g\ mL^{-1}}$.

Relation Between Molality and Mole Fraction

For a solution of a solute in a solvent of molar mass $M_{\mathrm{solvent}}$ (in $\mathrm{g\ mol^{-1}}$), the mole fraction of the solute is:

$x_{\mathrm{solute}} = \dfrac{m \times M_{\mathrm{solvent}}/1000}{1 + m \times M_{\mathrm{solvent}}/1000}$

For a $1\ \mathrm{m}$ aqueous solution, $M_{\mathrm{solvent}} = 18$, so $x_{\mathrm{solute}} = \dfrac{0.018}{1.018} \approx 0.0177$. This is the same as $1$ mol of solute in $1000/18 = 55.5$ mol of water.

Solved Examples on Molality Formula

Example 1: Molality From Number of Moles

Calculate the molality of a solution containing $0.5$ moles of $\ce{NaOH}$ dissolved in $2\ \mathrm{kg}$ of water.

Using the molality formula:

$m = \dfrac{n}{W}$

Substituting the given values:

$m = \dfrac{0.5}{2}$

$m = 0.25\ \mathrm{m}$

$\boxed{m = 0.25\ \mathrm{m}}$

Example 2: Molality From Mass of Solute

Find the molality of a solution prepared by dissolving $2.5\ \mathrm{g}$ of acetic acid ($\ce{CH3COOH}$) in $75\ \mathrm{g}$ of benzene.

Given: $w = 2.5\ \mathrm{g}$, $M_{\mathrm{molar}} = 60\ \mathrm{g\ mol^{-1}}$, $W_{\mathrm{g}} = 75\ \mathrm{g}$

Using:

$m = \dfrac{w}{M_{\mathrm{molar}}} \times \dfrac{1000}{W_{\mathrm{g}}}$

Therefore,

$m = \dfrac{2.5}{60} \times \dfrac{1000}{75}$

$m = 0.0417 \times 13.33$

$m \approx 0.556\ \mathrm{m}$

$\boxed{m \approx 0.556\ \mathrm{m}}$

Example 3: Molality of Sugar Solution

What is the molality of a solution containing $34.2\ \mathrm{g}$ of sugar $\ce{C12H22O11}$ dissolved in $500\ \mathrm{g}$ of water?

Given: $w = 34.2\ \mathrm{g}$, $M_{\mathrm{molar}} = 342\ \mathrm{g\ mol^{-1}}$, $W_{\mathrm{g}} = 500\ \mathrm{g}$

Using:

$m = \dfrac{w}{M_{\mathrm{molar}}} \times \dfrac{1000}{W_{\mathrm{g}}}$

Therefore,

$m = \dfrac{34.2}{342} \times \dfrac{1000}{500}$

$m = 0.1 \times 2$

$m = 0.2\ \mathrm{m}$

$\boxed{m = 0.2\ \mathrm{m}}$

Example 4: Molality From Molarity and Density

A solution has a molarity of $2\ \mathrm{M}$, density $1.2\ \mathrm{g\ mL^{-1}}$, and the solute is $\ce{NaCl}$ ($M_{\mathrm{molar}} = 58.5\ \mathrm{g\ mol^{-1}}$). Find the molality.

Given: $M = 2\ \mathrm{M}$, $d = 1.2\ \mathrm{g\ mL^{-1}}$, $M_{\mathrm{molar}} = 58.5\ \mathrm{g\ mol^{-1}}$

Using:

$m = \dfrac{1000M}{1000d - M \times M_{\mathrm{molar}}}$

Therefore,

$m = \dfrac{1000 \times 2}{1000 \times 1.2 - 2 \times 58.5}$

$m = \dfrac{2000}{1200 - 117}$

$m = \dfrac{2000}{1083}$

$m \approx 1.85\ \mathrm{mol\ kg^{-1}}$

$\boxed{m \approx 1.85\ \mathrm{m}}$

Example 5: Mass of Solvent Required for a Given Molality

Calculate the mass of water required to dissolve $9\ \mathrm{g}$ of glucose ($M_{\mathrm{molar}} = 180\ \mathrm{g\ mol^{-1}}$) to prepare a $0.25\ \mathrm{m}$ solution.

Given: $w = 9\ \mathrm{g}$, $M_{\mathrm{molar}} = 180\ \mathrm{g\ mol^{-1}}$, $m = 0.25\ \mathrm{m}$

Using:

$m = \dfrac{w}{M_{\mathrm{molar}}} \times \dfrac{1000}{W_{\mathrm{g}}}$

Rearranging:

$W_{\mathrm{g}} = \dfrac{w \times 1000}{M_{\mathrm{molar}} \times m}$

Substituting the values:

$W_{\mathrm{g}} = \dfrac{9 \times 1000}{180 \times 0.25}$

$W_{\mathrm{g}} = \dfrac{9000}{45}$

$W_{\mathrm{g}} = 200\ \mathrm{g}$

$\boxed{W_{\mathrm{g}} = 200\ \mathrm{g}}$

Example 6: Molarity From Molality

A $2\ \mathrm{m}$ aqueous solution of $\ce{NaCl}$ has a density of $1.2\ \mathrm{g\ mL^{-1}}$ ($M_{\mathrm{molar}} = 58.5\ \mathrm{g\ mol^{-1}}$). Find its molarity.

$M = \dfrac{1000\, m\, d}{1000 + m \times M_{\mathrm{molar}}} = \dfrac{1000 \times 2 \times 1.2}{1000 + 2 \times 58.5} = \dfrac{2400}{1117} \approx 2.15\ \mathrm{M}$

$\boxed{M \approx 2.15\ \mathrm{M}}$

Example 7: Boiling Point Elevation Using Molality

Find the elevation in the boiling point of a $0.5\ \mathrm{m}$ aqueous solution of a non-volatile solute. ($K_b$ of water $= 0.52\ \mathrm{K\ kg\ mol^{-1}}$)

$\Delta T_b = K_b \times m = 0.52 \times 0.5 = 0.26\ \mathrm{K}$

$\boxed{\Delta T_b = 0.26\ \mathrm{K}}$

Example 8: Molality of Pure Water

One kilogram of water has $\dfrac{1000}{18} = 55.5$ moles, so the molality of pure water is about $55.5\ \mathrm{mol\ kg^{-1}}$. This quick fact is often used in questions on dilute solutions.

Practice Questions on Molality Formula

1. Find the molality of a solution containing $5.85\ \mathrm{g}$ of $\ce{NaCl}$ ($58.5\ \mathrm{g\ mol^{-1}}$) in $250\ \mathrm{g}$ of water.
2. Find the molality when $18\ \mathrm{g}$ of glucose ($180\ \mathrm{g\ mol^{-1}}$) is dissolved in $200\ \mathrm{g}$ of water.
3. What mass of $\ce{KCl}$ ($74.5\ \mathrm{g\ mol^{-1}}$) is needed to prepare a $0.2\ \mathrm{m}$ solution in $500\ \mathrm{g}$ of water?

Answers: 1. $0.4\ \mathrm{m}$. 2. $0.5\ \mathrm{m}$. 3. $7.45\ \mathrm{g}$ (moles of $\ce{KCl}$ $= 0.2 \times 0.5 = 0.1$, and mass $= 0.1 \times 74.5$).

Common Mistakes While Using the Molality Formula

Advantages and Limitations of Molality

• Advantage: it does not change with temperature or pressure, because it uses mass and not volume.
• Advantage: it is the correct concentration term for colligative properties.
• Limitation: the solvent has to be weighed, so it is less convenient than measuring a volume when you prepare a solution in the lab.
• Limitation: it does not tell you directly how much solution to take, because it has no volume in it.

Important Formulas Related to Molality

Quick Revision: Molality Formula

The most important formula to remember is:

$\boxed{m = \dfrac{n}{W}}$

If mass is given:

$\boxed{m = \dfrac{w}{M_{\mathrm{molar}}} \times \dfrac{1000}{W_{\mathrm{g}}}}$

For conversion from molarity:

$\boxed{m = \dfrac{1000M}{1000d - M \times M_{\mathrm{molar}}}}$

This same set of formulas is equally important for NEET, so practise their application with the NEET Chapterwise PYQ, where colligative property numericals appear regularly.

Continue Your Mole Concept & Solutions Preparation

• Compare with the Molarity Formula guide to master the volume-based vs mass-based approach.
• Revise quickly with Mole Concept short notes before your exam.
• Practise more questions with solved numericals on Mole Concept, including molality-based problems.
• Attempt Mole Concept NEET previous year questions to understand how concentration concepts are tested.
• Cross-check your preparation with the NCERT Solutions for Class 11 Chemistry.