If the density of methanol is $0.793\ kg\ L^{-1}$, what is its volume needed for making 2.5 L of its 0.25 M solution?

If the density of methanol is 0.793kgL⁻¹, what is its volume needed for making 2.5 L of its 0.25 M solution?

Exercises · 1.12

0.793\ kg\ L^{-1}

Uttarakhand UbseTextbookSubjective· 3mImportance★★★★★est

22% · 20/89 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 — Molecular Mass Calculation

What is Molecular Mass? The Intuition

Imagine you're at a market buying apples. You don't weigh each apple individually — you put a dozen on a scale. The total weight tells you something about the apples, but it also depends on how many apples you have.

Atoms and molecules are unimaginably tiny. A single water molecule ( $H_2O$ ) weighs about $3 \times 10^{-23}$ grams. That number is useless for practical chemistry. So instead of working with individual molecules, chemists count them in huge fixed numbers — specifically, $6.022 \times 10^{23}$ molecules, which is called one mole.

Molecular mass is simply the mass of one mole of a substance, expressed in grams per mole (g/mol). It answers the question: "If I have $6.022 \times 10^{23}$ molecules of this compound, how much would they weigh on a lab balance?"

Note

The number $6.022 \times 10^{23}$ is Avogadro's constant. It's chosen so that the mass of one mole of carbon-12 atoms is exactly 12 grams — matching the atomic mass unit scale perfectly.

The Precise Definition

Molecular mass (also called molar mass) is the mass of one mole of a molecular substance. It is numerically equal to the sum of the atomic masses of all atoms in the molecule, expressed in g/mol.

For example:

Water ( $H_2O$ ): 2 hydrogen atoms + 1 oxygen atom
Atomic mass of H = 1.008 g/mol
Atomic mass of O = 16.00 g/mol
Molecular mass of $H_2O$ = $2(1.008) + 16.00 = 18.016$ g/mol

\text{Molecular mass} = \sum (\text{number of atoms of each element} \times \text{atomic mass of that element})

How to Calculate It — Step by Step

Let's take glucose, $C_6H_{12}O_6$ , as a worked example.

Step 1: Identify each element and its count

Carbon (C): 6 atoms
Hydrogen (H): 12 atoms
Oxygen (O): 6 atoms

Step 2: Look up atomic masses (from the periodic table)

C: 12.01 g/mol
H: 1.008 g/mol
O: 16.00 g/mol

Step 3: Multiply and add

\text{Molecular mass} = 6(12.01) + 12(1.008) + 6(16.00)

= 72.06 + 12.096 + 96.00

= 180.156 \text{ g/mol}

Tip

Always keep at least 2 decimal places from the periodic table. For exam problems, they usually give you atomic masses — use exactly what's provided.

Why This Matters

Molecular mass is the bridge between the microscopic world (atoms and molecules) and the macroscopic world (grams you can weigh). Once you know the molecular mass, you can:

Convert grams to moles: $\text{moles} = \frac{\text{mass in grams}}{\text{molecular mass}}$
Convert moles to grams: $\text{mass} = \text{moles} \times \text{molecular mass}$
Determine the number of molecules: $\text{molecules} = \text{moles} \times 6.022 \times 10^{23}$

Watch out

Do not confuse molecular mass with atomic mass. Atomic mass refers to a single element (like oxygen = 16.00 g/mol). Molecular mass refers to a compound (like $CO_2$ = 44.01 g/mol). Also, for ionic compounds like NaCl, we use formula mass (same calculation, but the substance isn't molecular).

Common Exam Pitfalls

Forgetting to multiply by the subscript. In $H_2SO_4$ , there are 2 hydrogens, not 1. …

Why this formula?

Stoichiometry & Mole Calculation: The "Why" Behind the Formula

Let's build this from the ground up — not as a list of formulas to memorise, but as a logical chain of reasoning.

1. The Core Question: What is a Mole?

A mole is simply a counting unit, like a dozen (12) or a gross (144). But instead of 12, a mole contains $6.022 \times 10^{23}$ particles (Avogadro's number, $N_A$ ).

Why this number?

It was chosen so that 1 mole of any substance has a mass in grams equal to its atomic/molecular mass in amu.

Example: 1 atom of carbon-12 has mass 12 amu.
1 mole of carbon-12 has mass 12 grams.

This is the bridge between the microscopic (atoms/molecules) and the macroscopic (grams we can weigh).

2. The Fundamental Relationship

The key formula is:

n = \frac{m}{M}

Where:

$n$ = number of moles
$m$ = mass of substance (in grams)
$M$ = molar mass (in g/mol)

Why does this work?

Think of it as a conversion factor:

If 1 mole of a substance weighs $M$ grams, then $m$ grams contains $\frac{m}{M}$ moles.

Derivation logic:

Molar mass $M$ tells you: "1 mol = $M$ g"
So the conversion factor is $\frac{1 \text{ mol}}{M \text{ g}}$
Multiply mass $m$ by this factor: $m \times \frac{1}{M} = \frac{m}{M}$ moles

3. Connecting to Number of Particles

n = \frac{N}{N_A}

Where:

$N$ = number of particles (atoms, molecules, ions)
$N_A = 6.022 \times 10^{23}$ particles/mol

Why?

1 mole = $N_A$ particles
So $N$ particles = $\frac{N}{N_A}$ moles

Combine both formulas:

\frac{m}{M} = \frac{N}{N_A}

This single equation ties mass, molar mass, number of particles, and Avogadro's number together.

4. The Gas Volume Connection (for gases at STP)

For gases only:

n = \frac{V}{22.4 \text{ L/mol}}

Why 22.4 L?

From the ideal gas law: $PV = nRT$

At STP (Standard Temperature and Pressure: 0°C, 1 atm):

$P = 1$ atm
$T = 273.15$ K
$R = 0.0821$ L·atm/(mol·K)

For $n = 1$ mole:

V = \frac{nRT}{P} = \frac{1 \times 0.0821 \times 273.15}{1} \approx 22.4 \text{ L}

So 1 mole of any ideal gas occupies 22.4 L at STP. This is a consequence of the gas laws, not a definition.

5. The Stoichiometry Chain: From One Substance to Another

In a balanced chemical equation like:

aA + bB \rightarrow cC + dD

The coefficients tell you the mole ratio:

\frac{\text{moles of A}}{\text{moles of B}} = \frac{a}{b}

Why this works: …

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

Moles needed $=0.25\times2.5=0.625$ mol; mass $=0.625\times32=20$ g; volume $=\dfrac{20}{0.793\ \text{g/mL}}\approx25.2$ mL.

Moles of methanol required.

n=M\times V=0.25\ \text{mol L}^{-1}\times2.5\ \text{L}=0.625\ \text{mol}.

Mass required. Molar mass of methanol $\mathrm{CH_3OH}=32\ \text{g mol}^{-1}$ :

$m=0.625\times32=20\ \text{g}.$ …

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

🧠 The Core Concept First

We need volume of pure methanol to prepare a dilute solution.

The steps are:

Find moles of methanol needed using molarity and final volume.
Convert moles to mass using molar mass.
Convert mass to volume using density.

Key formula:

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

✗ Common Mistake #1: Forgetting to convert density units

The error:

Density is given as $0.793\ kg\ L^{-1}$ , but molar mass is in $g\ mol^{-1}$ . Students often plug in $0.793$ directly without converting to $g\ L^{-1}$ .

How to avoid:

Always convert density to g L⁻¹ before using:

0.793\ kg\ L^{-1} = 793\ g\ L^{-1}

Why this matters:

Molar mass of methanol ( $CH_3OH$ ) = $32\ g\ mol^{-1}$ . If you use $0.793\ kg\ L^{-1}$ , your mass and volume units won’t cancel properly.

✗ Common Mistake #2: Confusing “volume of solution” with “volume of pure methanol”

The error:

Students think the 2.5 L is the volume of methanol needed.

Actually, 2.5 L is the final solution volume (methanol + water).

How to avoid:

Read carefully:

2.5 L = volume of solution (final diluted mixture)
We need volume of pure methanol (the solute) to prepare that solution.

✗ Common Mistake #3: Using wrong molar mass

The error:

Methanol is $CH_3OH$ , not $CH_4O$ or $CH_3O$ . Some students forget the H in the OH group.

How to avoid:

Write the formula clearly:

CH_3OH

Molar mass = $12 + (3 \times 1) + 16 + 1 = 32\ g\ mol^{-1}$

Double-check: C=12, H=4 (total), O=16 → 32 g/mol.

✗ Common Mistake #4: Unit mismatch in density conversion

The error:

Density is $0.793\ kg\ L^{-1}$ . Students multiply by 1000 but forget the unit — ending up with $793\ kg\ L^{-1}$ or $0.000793\ g\ L^{-1}$ .

How to avoid:

Use dimensional analysis:

0.793\ \frac{kg}{L} \times \frac{1000\ g}{1\ kg} = 793\ \frac{g}{L}

✓ Step-by-Step Correct Solution (for reference)

Step 1: Moles of methanol needed …

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

←Previous question1.11 — What is the concentration of sugar (C₁₂H₂₂O₁₁) in molL⁻¹ if its 20 g a…→Next question1.13 — Pressure is determined as force per unit area of the surface. The SI u…

Q.If the density of methanol is $0.793\ kg\ L^{-1}$ , what is its volume needed for making 2.5 L of its 0.25 M solution?

Concept understanding — Molecular Mass Calculation

What is Molecular Mass? The Intuition

The Precise Definition

How to Calculate It — Step by Step

Why This Matters

Common Exam Pitfalls

Why this formula?

Stoichiometry & Mole Calculation: The "Why" Behind the Formula

1. The Core Question: What is a Mole?

2. The Fundamental Relationship

Why does this work?

3. Connecting to Number of Particles

Why?

4. The Gas Volume Connection (for gases at STP)

Why 22.4 L?

5. The Stoichiometry Chain: From One Substance to Another

Why this works: …

Method: Molarity-Density Route (Mass → Moles → Volume)

Steps

🧠 The Core Concept First

✗ Common Mistake #1: Forgetting to convert density units

✗ Common Mistake #2: Confusing “volume of solution” with “volume of pure methanol”

✗ Common Mistake #3: Using wrong molar mass

✗ Common Mistake #4: Unit mismatch in density conversion

✓ Step-by-Step Correct Solution (for reference)

Q.If the density of methanol is 0.793 kg L−10.793\ kg\ L^{-1}0.793 kg L−1, what is its volume needed for making 2.5 L of its 0.25 M solution?

🎓Unlock everything free for 14 days

Q.If the density of methanol is $0.793\ kg\ L^{-1}$ , what is its volume needed for making 2.5 L of its 0.25 M solution?

Unlock everything free for 14 days