Molality Formula: Formula for Molality, Derivation & Solved Examples

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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, 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 — a property that makes it one of the most important concentration formulas in Class 11 and Class 12 Chemistry, especially for colligative property numericals like boiling point elevation and freezing point depression in JEE and NEET.

What Is Molality?

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

Therefore,

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

Where:

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.

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 formula for molality 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:

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

SI Unit and Symbol of Molality

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

Note: Molality is one of the few concentration terms whose unit, $\mathrm{mol,kg^{-1}}$, is already expressed directly in SI base units — this is why it's preferred for precise, temperature-sensitive experimental work.

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 — see the complete Molarity Formula guide for the volume-based approach and dilution equation.

Key Difference Between Molality and Molarity

The main difference is that molality uses the mass of solvent, whereas molarity uses the volume of solution.

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 colligative property laws — $\Delta T_b = K_b \times m$ and $\Delta T_f = K_f \times m$ — are defined using molality rather than molarity.

Relation Between Molarity and Molality

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

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

This relation is frequently asked in JEE Main and JEE Advanced, especially in problems where density is given instead of the mass of solvent directly.

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:

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

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

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

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

Once these solved patterns feel familiar, test your speed with eSaral's JEE Test Series, which includes dedicated Mole Concept and Solutions mock sections.

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:

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

Molality Formula at a Glance

This same set of formulas is equally important for NEET — practice their application with 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.

Want structured practice on every concentration formula, not just this one? Explore eSaral's JEE Course or NEET Course, both built around exactly these numerical patterns.