What will be the molarity of a solution, which contains 5.85 g of NaCl(s) per 500 mL? (i) 4mol L⁻¹ (ii) 20mol

NCERT Exemplar · Q3

Q. What will be the molarity of a solution, which contains 5.85 g of NaCl(s) per 500 mL?

(i)

4\ \text{mol L}^{-1}

(ii)

20\ \text{mol L}^{-1}

(iii)

0.2\ \text{mol L}^{-1}

(iv)

2\ \text{mol L}^{-1}

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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:

Colligative properties — properties like boiling point elevation and freezing point depression depend on the number of solute particles per mass of solvent, not per volume. Molality is the natural choice here.
Temperature-varying experiments — if you're working at different temperatures, molality keeps your concentration constant while molarity would drift.

Quick Comparison: Molarity vs Molality

Property	Molarity (M)	Molality (m)
Definition	moles solute / L solution	moles solute / kg solvent
Depends on temperature?	Yes (volume changes)	No (mass is constant)
Common unit	mol/L	mol/kg
Best used for	Room-temp reactions, titrations	Colligative properties, temperature studies

Final Takeaway

Molality is the concentration measure that stays honest when temperature changes. It's moles of solute per kilogram of solvent — and that's the whole story. Once you remember that the denominator is solvent mass, not solution volume, you've got it.

"Molality formula and calculation examples" and "molarity vs molality class 12 chemistry" are frequently searched terms, both grounded in the Solutions chapter of the NCERT/CBSE Class 12 Chemistry curriculum. Molality-based numericals are a near-guaranteed question type in board exams and JEE Main colligative-properties problems.

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?

Because molality requires solvent mass in kg, but we usually measure it in grams. The factor 1000 converts grams to kilograms:

1 \text{ kg} = 1000 \text{ g}

So if solvent mass is in grams, we multiply by 1000 to get the correct denominator in kg.

Common Mistake to Avoid

Do not use the mass of the solution (solute + solvent) in the denominator. The formula specifically requires mass of solvent only.

Example: If you dissolve 10 g NaCl in 90 g water, the solvent mass is 90 g, not 100 g.

Quick Check: Why This Matters in Exams

In problems involving:

Freezing point depression: $\Delta T_f = K_f \times m$
Boiling point elevation: $\Delta T_b = K_b \times m$

You must use molality, not molarity. The formula above is how you calculate $m$ from given masses.

Bottom line: Molality = moles of solute per kg of solvent. The ×1000 factor is just a unit conversion. The real conceptual leap is understanding why we use solvent mass — for temperature independence.

Molarity is moles of solute per litre of solution. For 5.85 g NaCl in 500 mL, moles = 0.1, volume in litres = 0.5, so molarity = 0.2 mol L⁻¹. The correct option is (iii).

Molarity is a measure of concentration: it tells you how many moles of solute are dissolved in one litre of the entire solution (not the solvent). The formula is:

\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in litres}}

The key here is to convert the given mass into moles, and the given volume into litres, then divide.

Find moles of NaCl. The molar mass of NaCl is $23 + 35.5 = 58.5\ \text{g mol}^{-1}$ . Given mass = 5.85 g.

\text{Moles} = \frac{5.85}{58.5} = 0.1\ \text{mol}

Convert volume to litres. Volume given = 500 mL.

500\ \text{mL} = 0.5\ \text{L}

Calculate molarity.

\text{Molarity} = \frac{0.1\ \text{mol}}{0.5\ \text{L}} = 0.2\ \text{mol L}^{-1}

Watch out

A common mistake is to forget to convert mL to L. If you used 500 directly, you'd get $0.1/500 = 0.0002$ , which is not among the options — but if you mistakenly used 500 as litres, you'd get $0.1/500 = 0.0002$ again, or worse, if you inverted, you'd get 5000. Always check: molarity of a dilute salt solution is small, around 0.1–1 M, not 4 or 20 M.

Tip

Notice that 5.85 g is exactly 0.1 times the molar mass (58.5 g). So you can think: "0.1 mole in half a litre" → double it to get moles per litre: $0.1 \times 2 = 0.2$ M. This mental shortcut saves time in exams.

✓Final answer

\boxed{0.2\ \text{mol L}^{-1}}

Common Mistakes in Molality/Molarity Calculation

This question tests molarity (not molality). The correct answer is (iii) $0.2\ \text{mol L}^{-1}$ .

Here are the most frequent errors students make:

1. Confusing Molarity with Molality

Mistake: Using mass of solvent instead of volume of solution.

Why it happens: The terms sound similar, and both start with "mol-".

How to avoid:

Molarity (M) = moles of solute per litre of solution
Molality (m) = moles of solute per kg of solvent
In this problem, volume is given (500 mL) → it's molarity.

2. Forgetting to Convert Volume to Litres

Mistake: Using 500 directly in the formula.

Example:

$M = \frac{5.85}{58.5 \times 500} = 0.0002$ (wrong)

How to avoid:

Always write the conversion step:

$500\ \text{mL} = 0.5\ \text{L}$

Then:

$M = \frac{\text{moles}}{\text{volume in L}} = \frac{0.1}{0.5} = 0.2\ \text{mol L}^{-1}$

3. Incorrect Molar Mass of NaCl

Mistake: Using Na = 23, Cl = 35.5 → but adding incorrectly (e.g., 23 + 35 = 58).

Result: Wrong mole count.

How to avoid:

Na = 23 g/mol, Cl = 35.5 g/mol
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Moles of NaCl = $\frac{5.85}{58.5} = 0.1\ \text{mol}$

4. Misreading the Question as Molality

Mistake: Assuming 500 mL is the solvent volume and calculating molality.

Example:

$m = \frac{0.1}{0.5} = 0.2\ \text{mol/kg}$ (coincidentally same number, but wrong concept)

How to avoid:

The question says "per 500 mL" — this is solution volume, not solvent.
For molality, you'd need mass of water (in kg), which is not given.

5. Arithmetic Slip in Final Division

Mistake: $0.1 \div 0.5 = 0.02$ or $2$ instead of $0.2$ .

How to avoid:

Write it as a fraction: $\frac{0.1}{0.5} = \frac{1}{5} = 0.2$
Or multiply numerator and denominator by 10: $\frac{1}{5} = 0.2$

Quick Checklist for Molarity Problems

Step	Action	Example
1	Find molar mass	NaCl = 58.5 g/mol
2	Convert mass to moles	$5.85 \div 58.5 = 0.1$ mol
3	Convert volume to litres	500 mL = 0.5 L
4	Apply formula	$M = 0.1 / 0.5 = 0.2$

Final answer: $\boxed{0.2\ \text{mol L}^{-1}}$ (Option (iii))

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

Showing the 12 most recent of 31 on this concept.

CBSE 2026Set 56/3/11 markMCQ
$2.0$
›Reveal solutionSolution
A 2.0 molal solution means 2 moles of solute in 1 kg of water. Convert the solvent mass to moles, then apply the mole fraction formula: $\chi_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}}$ . The answer is $\boxed{0.0347}$ .

Molality is defined as moles of solute per kilogram of solvent, not per kilogram of solution. This distinction matters because we need to count the moles of both solute and solvent separately to find the mole fraction.

When we say a solution is 2.0 molal, we're saying there are 2.0 moles of solute dissolved in exactly 1000 g (1 kg) of water. The mole fraction then asks: what fraction of the total number of particles (molecules) in the solution comes from the solute?

Let me work through the calculation systematically.

1. Identify what we know from "2.0 molal aqueous solution"

The molality $m = 2.0$ tells us:

Moles of solute: $n_{\text{solute}} = 2.0 \text{ mol}$

Mass of water (solvent): $1000 \text{ g}$

2. Convert the mass of water to moles

Water has a molar mass of $18 \text{ g/mol}$ , so:
$n_{\text{water}} = \frac{1000 \text{ g}}{18 \text{ g/mol}} = 55.56 \text{ mol}$
3. Calculate the total moles in the solution
$n_{\text{total}} = n_{\text{solute}} + n_{\text{water}} = 2.0 + 55.56 = 57.56 \text{ mol}$
4. Apply the mole fraction formula

The mole fraction of solute is:
$\chi_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{total}}} = \frac{2.0}{57.56} = 0.03474$
Rounding to three significant figures: $\chi_{\text{solute}} = 0.0347$

Watch out

A common mistake is to confuse molality (moles per kg of solvent) with molarity (moles per liter of solution). For molality, the denominator mass refers only to the solvent, which is why we can directly count 1000 g of water here.

Tip

For dilute aqueous solutions, you can use the quick approximation $\chi_{\text{solute}} \approx \frac{m}{55.56}$ where $m$ is the molality, since water contributes roughly 55.56 moles per kg. Here: $\frac{2.0}{55.56} \approx 0.036$ , close to our answer.

✓Final answer
$0.0347$
CBSE 2026Set A1 markMCQ
$Q. 34.2 g of sugar is present in 234.2 g of its aqueous solution. Then its molal concentration is$
(a) 0.1
(b) 0.5
(c) 5.5
(d) 55.0
›Reveal solutionSolution

Moles of sugar = 34.2/342 = 0.1; mass of solvent (water) = 234.2 - 34.2 = 200 g = 0.2 kg; molality = 0.1/0.2 = 0.5 m.

Molar mass of sugar (sucrose) = 342 g/mol.

Moles of sugar = 34.2/342 = 0.1 mol.

Mass of solution = 234.2 g, mass of sugar = 34.2 g, so mass of water = 234.2 - 34.2 = 200 g = 0.2 kg.

Molality = moles of solute / mass of solvent in kg = 0.1/0.2 = 0.5 mol/kg.

✓Final answer
$(b) 0.5.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. What will be the molarity of 30 ml of 0.5 M H2SO4 solution diluted to 50 ml?$
(a) 0.3 M
(b) 0.03 M
(c) 3 M
(d) 0.13 M
›Reveal solutionSolution

Dilution does not change the number of moles of solute, so M1V1 (before) = M2V2 (after).

Given: M1 = 0.5 M, V1 = 30 mL, final volume V2 = 50 mL.

Using the dilution law: M1V1 = M2V2

0.5 x 30 = M2 x 50

15 = 50 x M2

M2 = 15/50 = 0.3 M

✓Final answer
$(a) 0.3 M.$
CBSE 2026Set ANNUAL1 mark
$Q.Define mole fraction.$
›Reveal solutionSolution

Mole fraction expresses a component's amount relative to the total moles in a mixture, independent of temperature.

For a solution containing components with n1, n2, n3, ... moles, the mole fraction of component 1 is defined as:

x1 = n1 / (n1 + n2 + n3 + ...)

It is a dimensionless quantity, and the mole fractions of all components in a solution always sum to 1. Unlike molarity, mole fraction does not depend on temperature, since it is based purely on the number of moles, not volume.

✓Final answer
$Mole fraction = (moles of a given component) / (total moles of all components in the solution).$
CBSE 2026Set ANNUAL1 markMCQ
$Q. In an acid-base titrimetric analysis, the concentration of a sulphuric acid analyte is found to be 0.044 M. The strength of the acid in g/L is -$
(a) 0.44
(b) 4.31
(c) 2.15
(d) 44.00
›Reveal solutionSolution

Strength in g/L = molarity × molar mass = 0.044 × 98 ≈ 4.31 g/L, so option (B).

The strength of a solution in grams per litre is related to its molarity by

$\text{Strength (g/L)} = \text{Molarity (mol/L)} \times \text{Molar mass (g/mol)}.$

The molar mass of sulphuric acid $H_2SO_4 = 2(1) + 32 + 4(16) = 98\ g\,mol^{-1}$ .

$\text{Strength} = 0.044 \times 98 = 4.312 \approx 4.31\ g\,L^{-1}.$

✓Final answer
$(B) 4.31 g/L.$
CBSE 2025Set ANNUAL1 mark
$Q. Write the definition of molality.$
›Reveal solutionSolution

Molality expresses concentration as moles of solute per kilogram of SOLVENT (not solution), and unlike molarity it does not change with temperature.

Definition:

Molality (m) = (moles of solute) / (mass of solvent in kg)

Unit: mol kg^-1 (also written 'molal', symbol m)

Because molality is defined using the MASS of solvent (which does not change with temperature) rather than the VOLUME of solution (which expands/contracts with temperature), molality is a temperature-independent measure of concentration - unlike molarity, which does vary with temperature.

✓Final answer
$Molality = moles of solute per kilogram of solvent.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. What is the molarity of a solution with a mass of solute 10 kg mass and 100 litre volume?$
(a) 0.1 molar
(b) 1 molar
(c) 10 molar
(d) 100 molar
›Reveal solutionSolution

Molarity is defined as moles of solute per litre of solution; dividing the given amount of solute by the solution volume gives 0.1 M.

Molarity is defined as:

$M = \dfrac{\text{moles of solute}}{\text{volume of solution in litres}}$

Note: as printed, the question states the solute quantity as '10 kg mass'; for the arithmetic to match any of the given options (0.1, 1, 10, 100 M) the intended quantity is 10 moles of solute (a common wording/printing slip in this recurring question, where 'kg' should read 'mol') - with no molar mass given, that is the only value that lets the problem be solved from the stated data. Taking the solute amount as 10 mol:

$M = \dfrac{10\ \text{mol}}{100\ \text{L}} = 0.1\ \text{mol/L}$

✓Final answer
$(a) 0.1 molar.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. A solution contains 8 moles of solute and the mass of solvent is 4 kg. What is the molality of this solution?$
(a) 5 mol/kg
(b) 8 mol/kg
(c) 4 mol/kg
(d) 2 mol/kg
›Reveal solutionSolution
Molality is moles of solute per kilogram of solvent (not solution); here it works out to 8/4 = 2 mol/kg.

Molality (m) is defined as:

m = (moles of solute) / (mass of solvent in kg)

Given:

moles of solute = 8 mol

mass of solvent = 4 kg

m = 8 mol / 4 kg = 2 mol/kg

Note the common trap here: molality uses the mass of the SOLVENT, not the total solution mass or volume (that would be molarity or a different quantity) — a frequent point of confusion tested in this exact question style.

✓Final answer
$(d) 2 mol/kg.$
CBSE 2025Set ANNUAL1 mark
$Q. Define molality.$
›Reveal solutionSolution

Molality is defined as moles of solute per kilogram of solvent; unlike molarity, it does not depend on temperature since it is based on mass, not volume.

Molality (denoted m) of a solution is defined as the number of moles of solute dissolved in one kilogram (1000 g) of solvent:

Molality (m) = (number of moles of solute) / (mass of solvent in kg)

Unit: mol kg^-1 (also written as 'm', e.g., a '1 molal' or '1 m' solution).

Unlike molarity (which is based on the volume of the solution and therefore varies slightly with temperature, since volume expands/contracts with temperature), molality is based on the mass of the solvent, which does not change with temperature — so molality is a temperature-independent way of expressing concentration.

✓Final answer
$Molality is the number of moles of solute per kilogram of solvent: m = moles of solute / mass of solvent (in kg).$
CBSE 2024Set A11 mark
$Q.The number of moles of solute present in one kilogram of the solvent is called \rule{2cm}{0.4pt}.$
›Reveal solutionSolution

Moles of solute per kilogram of solvent defines molality.

Molality ( $m$ ) is a concentration term that depends only on the mass of solvent (and is therefore temperature-independent):

$m = \dfrac{\text{moles of solute}}{\text{mass of solvent in kg}} \;\; (\text{mol kg}^{-1})$

Since the definition specifies moles of solute in one kilogram of solvent, the quantity described is molality.

✓Final answer
$molality$
CBSE 2024Set ANNUAL1 markMCQ
$Q. Which of the following has no unit?$
(a) Molarity
(b) Molality
(c) Normality
(d) Molar Fraction
›Reveal solutionSolution

Mole fraction is a dimensionless ratio, so it has no unit.

Molarity has units of $mol\,L^{-1}$ , molality has units of $mol\,kg^{-1}$ , and normality has units of $eq\,L^{-1}$ . Mole (molar) fraction is defined as the ratio of moles of one component to the total moles of all components in the solution, $x_i = \dfrac{n_i}{\sum n}$ — since it is a ratio of like quantities, it is a pure number and carries no unit.

✓Final answer
$(iv) Molar Fraction$
CBSE 2024Set ANNUAL1 mark
$Q. Write down the formula of Molarity.$
›Reveal solutionSolution

Molarity is the number of moles of solute dissolved per litre of solution.

Molarity is one of the most widely used units of concentration. It is defined as the number of moles of solute dissolved in one litre (one cubic decimetre) of solution:

$M = \frac{\text{moles of solute } (n)}{\text{volume of solution in litres } (V)}$

Equivalently, in terms of mass, $M = \dfrac{w \times 1000}{M_{r} \times V(mL)}$ , where $w$ is the mass of solute in grams and $M_r$ is its molar mass.

✓Final answer
$M = \dfrac{n}{V(L)}\ mol\,L^{-1}$

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Q.What will be the molarity of a solution, which contains 5.85 g of NaCl(s) per 500 mL?

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?

Quick Comparison: Molarity vs Molality

Final Takeaway

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?

Common Mistake to Avoid

Quick Check: Why This Matters in Exams

Method: Molarity Formula (Direct Substitution)

Steps

Common Mistakes in Molality/Molarity Calculation

1. Confusing Molarity with Molality

2. Forgetting to Convert Volume to Litres

3. Incorrect Molar Mass of NaCl

4. Misreading the Question as Molality

5. Arithmetic Slip in Final Division

Quick Checklist for Molarity Problems

Q.What will be the molarity of a solution, which contains 5.85 g of NaCl(s) per 500 mL?

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