The density of 3 molal solution of NaOH is $1.110\ \text{g mL}^{-1}$. Calculate the molarity of the solution.

The density of 3 molal solution of NaOH is 1.110g mL⁻¹. Calculate the molarity of the solution.

NCERT Exemplar · Q32

1.110\ \text{g mL}^{-1}

Kerala DhseShort· 3mImportance★★★★★

85% · 76/89 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Molality Calculation

Molality: The Concentration That Ignores Temperature

Imagine you're making a cup of sweet tea. You add sugar to hot water, stir, and taste. If you let the tea cool to room temperature, the amount of sugar hasn't changed — but the volume of the liquid has shrunk slightly. If you measured concentration as "grams of sugar per litre of solution," that number would change just because the temperature changed. That's annoying if you're a chemist who needs a reliable, temperature-independent way to describe how much solute is present.

Molality was invented to solve exactly this problem.

The Intuition

Instead of measuring the volume of the solution (which expands and contracts with temperature), molality measures the mass of the solvent. Mass doesn't change with temperature. So molality gives you a concentration that stays the same whether your solution is hot or cold.

Think of it this way:

Molarity = moles of solute per litre of solution (temperature-sensitive)
Molality = moles of solute per kilogram of solvent (temperature-independent)

The solvent is the substance doing the dissolving — usually water. The solute is what gets dissolved — sugar, salt, etc.

The Precise Definition

\text{Molality } (m) = \frac{\text{moles of solute}}{\text{kilograms of solvent}}

The symbol for molality is a lowercase m (not to be confused with M for molarity).

Key points to remember:

The denominator is solvent mass, not solution mass
The unit is mol/kg (often written as simply "m")
It is independent of temperature because mass doesn't change with temperature

Worked Example

Problem: 36 g of glucose ( $C_6H_{12}O_6$ , molar mass = 180 g/mol) is dissolved in 500 g of water. Calculate the molality of the solution.

Step 1: Find moles of solute

\text{Moles of glucose} = \frac{36 \text{ g}}{180 \text{ g/mol}} = 0.2 \text{ mol}

Step 2: Convert solvent mass to kilograms

500 \text{ g} = 0.5 \text{ kg}

Step 3: Apply the formula

m = \frac{0.2 \text{ mol}}{0.5 \text{ kg}} = 0.4 \text{ m}

Note

The answer is 0.4 m (or 0.4 mol/kg). Notice we used the mass of water (500 g), not the mass of the solution (which would be 536 g).

Common Mistake to Avoid

Watch out

Do not use the mass of the solution in the denominator. The formula specifically asks for the mass of the solvent alone. If the problem gives you the total mass of the solution, subtract the mass of the solute to find the solvent mass.

When Do You Use Molality?

Molality is the star in two important situations: …

Why this formula?

Molality Calculation: Why the Formula Works

Molality is a measure of concentration that is temperature-independent — this is its key advantage over molarity. Let's understand why the formula takes the form it does.

The Definition First

Molality ( $m$ ) is defined as:

m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}

The unit is mol/kg, often written as m (e.g., 0.5 m glucose solution).

Why Mass of Solvent, Not Solution?

This is the critical conceptual point.

The Reasoning

Molarity uses volume of solution → volume changes with temperature (expansion/contraction). So molarity changes with temperature.
Molality uses mass of solvent → mass is invariant with temperature. So molality remains constant regardless of temperature changes.

Key insight: By using the solvent's mass (not the solution's volume), we eliminate temperature dependence. This is why molality is preferred for colligative properties (boiling point elevation, freezing point depression) — these properties depend on the number of solute particles, not on temperature.

Deriving the Formula Step-by-Step

Step 1: Moles of Solute

If you have $w_{\text{solute}}$ grams of solute with molar mass $M_{\text{solute}}$ (g/mol):

\text{moles of solute} = \frac{w_{\text{solute}}}{M_{\text{solute}}}

Step 2: Mass of Solvent in kg

If the solvent mass is $W_{\text{solvent}}$ grams:

\text{mass of solvent in kg} = \frac{W_{\text{solvent}}}{1000}

Step 3: Putting It Together

m = \frac{\frac{w_{\text{solute}}}{M_{\text{solute}}}}{\frac{W_{\text{solvent}}}{1000}}

Simplifying:

m = \frac{w_{\text{solute}} \times 1000}{M_{\text{solute}} \times W_{\text{solvent}}}

The Final Formula (Exam-Ready)

\boxed{m = \frac{w_{\text{solute}} \times 1000}{M_{\text{solute}} \times W_{\text{solvent}}}}

Where:

$w_{\text{solute}}$ = mass of solute in grams
$M_{\text{solute}}$ = molar mass of solute in g/mol
$W_{\text{solvent}}$ = mass of solvent in grams

Why the ×1000 Factor? …

Start your 14-day free trial to unlock the full solution →

Take $1$ kg of solvent (holding $3$ mol $NaOH$ ), find the solution mass, convert to volume with the given density, then divide moles by volume: molarity $= \mathbf{2.97\ M}$ .

1. Interpret 3 molal

A $3$ molal solution has $3$ mol $NaOH$ dissolved in $1$ kg $=1000$ g of water.

2. Mass of the solution

Molar mass of $NaOH = 23+16+1 = 40\ \text{g mol}^{-1}$ , so mass of $NaOH = 3\times40 = 120$ g.

\text{mass of solution}=1000+120=1120\ \text{g}

3. Volume of the solution

Using $\text{density}=1.110\ \text{g mL}^{-1}$ :

V=\frac{1120\ \text{g}}{1.110\ \text{g mL}^{-1}}=1009.0\ \text{mL}=1.009\ \text{L}

4. Molarity …

Start your 14-day free trial to unlock the full solution →

Common Mistakes in Molality → Molarity Calculation

Students often struggle with this conversion because it requires careful handling of mass, volume, and the distinction between molality and molarity. Here are the most frequent errors and how to avoid them.

Mistake 1: Confusing molality with molarity

The error:

Students treat 3 molal as 3 moles per litre of solution, which is actually molarity.

Why it happens:

Both terms sound similar and both use "moles of solute," but:

Molality ( $m$ ) = moles of solute per kg of solvent
Molarity ( $M$ ) = moles of solute per L of solution

How to avoid:

Always write the definition before starting:

$m = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}}$
$M = \frac{\text{moles of solute}}{\text{volume of solution (in L)}}$

Mistake 2: Using density to directly convert molality to molarity

The error:

Students multiply: $3 \times 1.110 = 3.33\ \text{M}$ — this is wrong.

Why it happens:

Density is mass/volume, but molality uses solvent mass, not solution mass.

How to avoid:

Remember the stepwise logic:

Assume 1 kg of solvent (this is the standard starting point for molality)
Find moles of solute from molality
Find mass of solute
Find total mass of solution = mass of solvent + mass of solute
Use density to find volume of solution
Finally, molarity = moles of solute / volume of solution (in L)

Mistake 3: Forgetting to convert units

The error:

Using density in g/mL but volume in litres without conversion.

Example of wrong work:

Volume = $\frac{\text{mass of solution (g)}}{\text{density (g/mL)}}$ → then using this volume directly in litres.

How to avoid:

Always convert mL to L:

$1\ \text{mL} = 10^{-3}\ \text{L}$
So if volume comes out as $V\ \text{mL}$ , then $V\ \text{mL} = \frac{V}{1000}\ \text{L}$

Mistake 4: Using wrong molar mass or forgetting it entirely

The error:

Students skip finding the mass of NaOH, or use an incorrect molar mass.

Why it happens:

They forget that to find the mass of solute from moles, you must know the molar mass.

How to avoid:

Molar mass of NaOH = $23 + 16 + 1 = 40\ \text{g/mol}$ …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KEAM 2026Set eng-2026-04174 marksMCQ
$M_2$
›Reveal solutionSolution

Convert molarity to molality by finding the mass of solvent in 1 L of solution; the result is $m = \dfrac{1000c}{1000\rho - cM_2}$ , identical to option (A).

Take exactly 1 litre ( $1000$ cm $^3$ ) of solution.

Mass of solution $= \text{volume} \times \text{density} = 1000\rho$ g.

Moles of solute in 1 L $= c$ (definition of molarity), so mass of solute $= cM_2$ g.

Mass of solvent $= (1000\rho - cM_2)$ g $= \dfrac{1000\rho - cM_2}{1000}$ kg. …
Start your 14-day free trial to see the full solution →
KEAM 2026Set eng-2026-04194 marksMCQ
$^{-1}$
›Reveal solutionSolution

A $0.5\,m$ solution in $100\,g$ water needs $0.05\,mol$ ( $3\,g$ ) of ethanoic acid.

Molality $=\dfrac{\text{moles of solute}}{\text{kg of solvent}}$ . For $0.5\,m$ in $100\,g=0.100\,kg$ water:

$n=0.5\times0.100=0.05\,mol.$ …
Start your 14-day free trial to see the full solution →
KEAM 2026Set eng-2026-04224 marksMCQ
$mL$
›Reveal solutionSolution

Moles $=$ mass/MW $=3.0/30=0.1$ mol in $0.25$ L, so molarity $=0.4$ M.

$n=\frac{3.0\ \text{g}}{30\ \text{g mol}^{-1}}=0.1\ \text{mol}$ …
Start your 14-day free trial to see the full solution →
KEAM 2026Set pha-2026-0418F4 marksMCQ
$Q. The molarity of an aqueous solution containing 0.4g of NaOH (molar mass=40g/mol) in 250mL of a solution is (A) 0.04M (B) 0.02M (C) 0.20M (D) 0.40M (E) 0.08M$
›Reveal solutionSolution

0.4 g NaOH is 0.01 mol; in 0.250 L that is 0.04 M.

Reasoning

Moles of NaOH $=\dfrac{0.4\ \text{g}}{40\ \text{g mol}^{-1}}=0.01$ mol.

Volume $=250\ \text{mL}=0.250$ L. …
Start your 14-day free trial to see the full solution →
KEAM 2025Set eng-2025-04234 marksMCQ
$1.86 \ g \ mL^{-1}$
›Reveal solutionSolution
The molality of the 3 M solution is 2 m.

Concept and Intuition

Molarity is per litre of solution; molality is per kilogram of solvent. Using the density to get the mass of 1 L of solution, subtracting the solute mass gives the solvent mass, from which molality follows.

Step-by-Step Solution

Mass of 1 L solution = 1000 mL * 1.86 g/mL = 1860 g.

Solute in 1 L (3 mol) = 3 * 120 = 360 g.

Solvent mass = 1860 - 360 = 1500 g = 1.5 kg. …
Start your 14-day free trial to see the full solution →
KEAM 2024Set eng-2024-06084 marksMCQ
$^{-1}$
›Reveal solutionSolution

Molality = moles of solute per kg of solvent. Here $1$ mol urea in $0.2$ kg water gives $5$ m.

The number of moles of urea is
$n=\frac{60\ \text{g}}{60\ \text{g mol}^{-1}}=1\ \text{mol}.$
The solvent (water) mass is the total solution mass minus the solute mass:

$260-60=200\ \text{g}=0.2\ \text{kg}.$ …
Start your 14-day free trial to see the full solution →
KEAM 2024Set eng-2024-06084 marksMCQ
$40\ g\ mol^{-1}$
›Reveal solutionSolution

Molarity = moles of solute per litre of solution: $0.2$ mol in $0.25$ L gives $0.8$ M.

The number of moles of NaOH is
$n=\frac{8\ \text{g}}{40\ \text{g mol}^{-1}}=0.2\ \text{mol}.$
The solution volume is $250\ \text{mL}=0.25\ \text{L}$ , so the molarity is …
Start your 14-day free trial to see the full solution →
KEAM 2024Set eng-2024-06094 marksMCQ
$^{-1}$
›Reveal solutionSolution

$6\ \mathrm{g}$ NaOH is $0.15\ \mathrm{mol}$ ; in $0.6\ \mathrm{L}$ this is $0.25\ \mathrm{M}$ .

Moles of NaOH:
$n = \dfrac{6}{40} = 0.15\ \mathrm{mol}.$
Molarity: …
Start your 14-day free trial to see the full solution →
KEAM 2024Set eng-2024-06094 marksMCQ
$^{-1}$
›Reveal solutionSolution

$3\ \mathrm{L}$ of $0.25\ \mathrm{M}$ needs $0.75\ \mathrm{mol}=45\ \mathrm{g}$ ethanol; at density $360\ \mathrm{g\,L^{-1}}$ that is $125\ \mathrm{mL}$ .

Moles required:
$n = M\times V = 0.25 \times 3 = 0.75\ \mathrm{mol}.$
Mass:

$m = n\times M_{molar} = 0.75 \times 60 = 45\ \mathrm{g}.$ …
Start your 14-day free trial to see the full solution →

←Previous questionQ31 — Hydrogen gas is prepared in the laboratory by reacting dilute HCl with…→Next questionQ33 — Volume of a solution changes with change in temperature, then, will th…

Q.The density of 3 molal solution of NaOH is $1.110\ \text{g mL}^{-1}$ . Calculate the molarity of the solution.

Concept understanding — Molality Calculation

Molality: The Concentration That Ignores Temperature

The Intuition

The Precise Definition

Worked Example

Common Mistake to Avoid

When Do You Use Molality?

Why this formula?

Molality Calculation: Why the Formula Works

The Definition First

Why Mass of Solvent, Not Solution?

The Reasoning

Deriving the Formula Step-by-Step

Step 1: Moles of Solute

Step 2: Mass of Solvent in kg

Step 3: Putting It Together

The Final Formula (Exam-Ready)

Why the ×1000 Factor? …

Method: Density–Molality to Molarity Conversion

Steps

Common Mistakes in Molality → Molarity Calculation

Mistake 1: Confusing molality with molarity

Mistake 2: Using density to directly convert molality to molarity

Mistake 3: Forgetting to convert units

Mistake 4: Using wrong molar mass or forgetting it entirely

Concept and Intuition

Step-by-Step Solution

Q.The density of 3 molal solution of NaOH is 1.110 g mL−11.110\ \text{g mL}^{-1}1.110 g mL−1. Calculate the molarity of the solution.

🎓Unlock everything free for 14 days

Q.The density of 3 molal solution of NaOH is $1.110\ \text{g mL}^{-1}$ . Calculate the molarity of the solution.

Unlock everything free for 14 days