Calculate the amount of benzoic acid ($C_6H_5COOH$) required for preparing 250 mL of 0.15 M solution in methanol.

> [!TLDR] > The key idea is to use molarity ($M = \frac{n}{V}$) to find moles of solute, then convert moles to mass via molar mass. For 250 mL of 0.15 M benzoic acid in methanol, the required mass is **4.58 g**. ## Why This Approach Works Molarity tells us the number of moles of solute dissolved in **one litre** of solution. When you know the volume you want to prepare, you can scale the moles proportionally. Once you have the moles, multiplying by the molar mass of benzoic acid gives the mass you need to weigh out. The solvent (methanol here) doesn't affect the calculation — it just carries the solute. The formula is straightforward: > [!FORMULA] > $$M = \frac{n}{V} \quad \Rightarrow \quad n = M \times V$$ > Then: $\text{mass} = n \times \text{molar mass}$ But watch out: volume must be in **litres**, not millilitres. That's the most common slip. --- ## Step-by-Step Calculation **1. Write down what's given.** - Molarity, $M = 0.15\ \text{mol L}^{-1}$ - Volume, $V = 250\ \text{mL} = 0.250\ \text{L}$ (convert by dividing by 1000) - Solute: benzoic acid, $C_6H_5COOH$ **2. Find the molar mass of benzoic acid.** Benzoic acid has the formula $C_7H_6O_2$ (since $C_6H_5COOH$ has 7 carbons, 6 hydrogens, 2 oxygens). Atomic masses (approx.): - Carbon: $12.0\ \text{g mol}^{-1}$ - Hydrogen: $1.0\ \text{g mol}^{-1}$ - Oxygen: $16.0\ \text{g mol}^{-1}$ So: $$ \text{Molar mass} = (7 \times 12.0) + (6 \times 1.0) + (2 \times 16.0) = 84.0 + 6.0 + 32.0 = 122.0\ \text{g mol}^{-1} $$ > [!TIP] > You can also think of benzoic acid as $C_6H_5COOH$: $C_6H_5$ (77 g/mol) + $COOH$ (45 g/mol) = 122 g/mol. Quick mental check. **3. Calculate the moles of benzoic acid needed.** …

Calculate the amount of benzoic acid (C₆H₅COOH) required for preparing 250 mL of 0.15 M solution in methanol.

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C_6H_5COOH

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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: …

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? …

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The key idea is to use molarity ( $M = \frac{n}{V}$ ) to find moles of solute, then convert moles to mass via molar mass. For 250 mL of 0.15 M benzoic acid in methanol, the required mass is 4.58 g.

Why This Approach Works

Molarity tells us the number of moles of solute dissolved in one litre of solution. When you know the volume you want to prepare, you can scale the moles proportionally. Once you have the moles, multiplying by the molar mass of benzoic acid gives the mass you need to weigh out. The solvent (methanol here) doesn't affect the calculation — it just carries the solute.

The formula is straightforward:

M = \frac{n}{V} \quad \Rightarrow \quad n = M \times V

But watch out: volume must be in litres, not millilitres. That's the most common slip.

Step-by-Step Calculation

1. Write down what's given.

Molarity, $M = 0.15\ \text{mol L}^{-1}$
Volume, $V = 250\ \text{mL} = 0.250\ \text{L}$ (convert by dividing by 1000)
Solute: benzoic acid, $C_6H_5COOH$

2. Find the molar mass of benzoic acid.

Benzoic acid has the formula $C_7H_6O_2$ (since $C_6H_5COOH$ has 7 carbons, 6 hydrogens, 2 oxygens).

Atomic masses (approx.):

Carbon: $12.0\ \text{g mol}^{-1}$
Hydrogen: $1.0\ \text{g mol}^{-1}$
Oxygen: $16.0\ \text{g mol}^{-1}$

So:

\text{Molar mass} = (7 \times 12.0) + (6 \times 1.0) + (2 \times 16.0) = 84.0 + 6.0 + 32.0 = 122.0\ \text{g mol}^{-1}

Tip

You can also think of benzoic acid as $C_6H_5COOH$ : $C_6H_5$ (77 g/mol) + $COOH$ (45 g/mol) = 122 g/mol. Quick mental check.

3. Calculate the moles of benzoic acid needed. …

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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:

$\chi_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{total}}} = \frac{2.0}{57.56} = 0.03474$ …
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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. …

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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.

…

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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 + ...)

…

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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

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

…

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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)

…

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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 n …

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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

…
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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).

…

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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):

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

…

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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:

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

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Q.Calculate the amount of benzoic acid ( $C_6H_5COOH$ ) required for preparing 250 mL of 0.15 M solution in methanol.

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? …

Why This Approach Works

Step-by-Step Calculation

Method: Molarity-to-Mass Conversion (via Moles)

Steps

🧪 Common Mistakes in Molality Calculation (Benzoic Acid in Methanol)

✓ The Core Idea

✗ Mistake #1: Confusing Molarity with Molality

✗ Mistake #2: Forgetting to Convert mL to L

✗ Mistake #3: Using Wrong Molar Mass

Q.Calculate the amount of benzoic acid (C6H5COOHC_6H_5COOHC6​H5​COOH) required for preparing 250 mL of 0.15 M solution in methanol.

Why This Approach Works

Step-by-Step Calculation

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Q.Calculate the amount of benzoic acid ( $C_6H_5COOH$ ) required for preparing 250 mL of 0.15 M solution in methanol.

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