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.

Exercises · 1.30

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.

TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
$0.892\,N$
›Reveal solutionSolution

" $20$ volume" means $1\ \text{L}$ of solution releases $20\ \text{L}$ of O $_2$ at STP; using $N=\dfrac{\text{volume strength}}{5.6}$ gives $N=\dfrac{20}{5.6}=3.57\ N$ - option (D).

Meaning of volume strength. A " $20$ volume" H $_2$ O $_2$ solution liberates $20\ \text{L}$ of O $_2$ (at STP) per litre of solution on decomposition:
$2\,\text{H}_2\text{O}_2\rightarrow2\,\text{H}_2\text{O}+\text{O}_2$
Step 1 - Moles of O $_2$ per litre.
$n_{\text{O}_2}=\frac{20}{22.4}=0.893\ \text{mol}$
Step 2 - Moles and equivalents of H $_2$ O $_2$ . …
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TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
$_3$
›Reveal solutionSolution
The key is to use the balanced chemical equation and stoichiometry to find the mass of pure HCl needed, then convert to the mass of the 50% w/w solution. The required mass is 292 g, so option (B) is correct.

Concept & Intuition

This problem is about reacting hydrochloric acid with calcium carbonate. The reaction is a classic acid–carbonate neutralization:
$\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O}$
We are given a 50% w/w solution — meaning 50 g of pure HCl per 100 g of solution. So the actual mass of solution needed will be double the mass of pure HCl required. The trap is forgetting to account for the dilution and just picking the mass of pure HCl.

Step-by-step solution

Write the balanced equation

$\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O}$
This tells us: 1 mole of CaCO₃ reacts with 2 moles of HCl.

Find moles of CaCO₃ Molar mass of CaCO₃ = 40 (Ca) + 12 (C) + 3×16 (O) = 100 g/mol. Given mass = 200 g.

$\text{Moles of CaCO}_3 = \frac{200}{100} = 2 \text{ mol}$

Find moles of HCl needed

From the equation: 1 mol CaCO₃ needs 2 mol HCl.

So 2 mol CaCO₃ need $2 \times 2 = 4$ mol HCl.

Find mass of pure HCl required

Molar mass of HCl = 1 + 35.5 = 36.5 g/mol.

$\text{Mass of pure HCl} = 4 \times 36.5 = 146 \text{ g}$ …
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TG EAPCET 2021Set ap-2021-08-10-AN1 markMCQ
$\mathrm{H_2O_2}$
›Reveal solutionSolution
The "volume strength" of hydrogen peroxide is the volume of $\mathrm{O_2}$ (at STP) released per volume of solution. For a 50-volume solution the strength (w/v) is about 15%, so the correct option is (D).

Concept & Intuition

"Volume strength" (e.g. "10 volume," "50 volume") labels a hydrogen peroxide solution by the volume of oxygen it releases: 1 mL of the solution produces that many mL of $\mathrm{O_2}$ at STP on decomposition. The decomposition is
$2\mathrm{H_2O_2} \rightarrow 2\mathrm{H_2O} + \mathrm{O_2}$
To convert volume strength into a percentage (g of $\mathrm{H_2O_2}$ per 100 mL of solution), use the molar volume at STP (22.4 L/mol) and the molar mass of $\mathrm{H_2O_2}$ (34 g/mol).

Step-by-step reasoning

Interpret "50 volume"

1 mL of solution yields 50 mL of $\mathrm{O_2}$ at STP, so 1 L of solution yields $50 \times 1000 = 50\,000$ mL = 50 L of $\mathrm{O_2}$ .

Moles of $\mathrm{O_2}$

$n_{\mathrm{O_2}} = \frac{50}{22.4} \approx 2.232\ \text{mol}$

Moles of $\mathrm{H_2O_2}$

From the 2:1 ratio, $n_{\mathrm{H_2O_2}} = 2 \times 2.232 = 4.464$ mol.

Mass of $\mathrm{H_2O_2}$ per litre

With molar mass 34 g/mol,

$m = 4.464 \times 34 \approx 151.8\ \text{g per litre}$

Express as a percentage (w/v) …
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TG EAPCET 2021Set ap-2021-08-10-FN1 markMCQ
$\mathrm{H_2O_2}$
›Reveal solutionSolution
"Volume strength" of $\mathrm{H_2O_2}$ is the volume of $\mathrm{O_2}$ (at STP) released per volume of solution. For a 50-volume solution, converting through the decomposition $2\mathrm{H_2O_2} \to 2\mathrm{H_2O} + \mathrm{O_2}$ gives a strength of about 15%, so the correct option is (D).

Why this approach works

"Volume strength" tells you how many millilitres of oxygen gas (at STP) one millilitre of the solution releases on decomposition. So a "50 volume" solution means 1 mL of solution yields 50 mL of $\mathrm{O_2}$ .

To convert this into a percentage by mass (w/v), we:

find the mass of $\mathrm{H_2O_2}$ that produces that volume of oxygen, using the decomposition stoichiometry;

relate that mass to the mass of solution (density approx 1 g/mL for dilute solutions).

The bridge is the balanced equation:
$2\mathrm{H_2O_2} \rightarrow 2\mathrm{H_2O} + \mathrm{O_2}$
so 2 moles of $\mathrm{H_2O_2}$ give 1 mole of $\mathrm{O_2}$ .

Step-by-step reasoning

Moles of oxygen from the given volume At STP, 1 mole of gas occupies 22400 mL. For 50 mL of $\mathrm{O_2}$ (from 1 mL of solution):

$n_{\mathrm{O_2}} = \frac{50}{22400} = \frac{1}{448}\ \text{mol}$

Moles of $\mathrm{H_2O_2}$ that produce this oxygen

From the 2:1 ratio, $n_{\mathrm{H_2O_2}} = 2 \times \frac{1}{448} = \frac{1}{224}\ \text{mol}$ .

Mass of $\mathrm{H_2O_2}$

Molar mass of $\mathrm{H_2O_2} = 34$ g/mol, so

$m = \frac{1}{224}\times 34 = \frac{34}{224} \approx 0.152\ \text{g}$ …
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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

Why this approach works

Step-by-step reasoning

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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