Q.The unit of length convenient on the atomic scale is known as an angstrom and is denoted by Å: 1 A˚=10−10 m. The size of a hydrogen atom is about 0.5 A˚. What is the total atomic volume in m3 of a mole of hydrogen atoms?
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Atomic Volume: Meaning and Calculation
Atoms are mostly empty space — a tiny dense nucleus wrapped in a fuzzy electron cloud. So "atomic volume" does not mean the volume of a solid ball; it means the average space one atom occupies when many atoms are packed together in a solid or liquid.
Think of a crowded hall: to find the space per person you divide the hall's volume by the number of people. Atomic volume does exactly that for atoms.
Definition
Atomic volume is the volume occupied by one mole of atoms of an element in its solid or liquid state. It is found from the element's molar mass and density:
Vatomic=ρM
- Vatomic = atomic volume (cm³/mol)
- M = molar mass (g/mol)
- ρ = density (g/cm³)
This gives the volume per mole. Dividing by Avogadro's number gives the space per single atom:
Vone atom=NAVatomic,NA=6.022×1023 mol−1
Worked example — aluminium
Molar mass M=26.98 g/mol, density ρ=2.70 g/cm³:
Vatomic=2.7026.98=9.99 cm3/mol
So one mole of Al atoms occupies about 10 cm³. Per atom:
Vone Al atom=6.022×10239.99=1.66×10−23 cm3
Periodic trends
- Down a group: atomic volume increases — more electron shells make atoms larger.
- Across a period: it generally decreases — rising nuclear charge pulls the electrons in tighter.
- Allotropes differ: diamond is denser than graphite, so diamond's atomic volume is smaller, though both are carbon.
These trends help explain why alkali metals (large atomic volume) are soft and reactive, while transition metals (smaller atomic volume) are hard and dense. …
Why this formula?
Atomic Volume Calculation: Understanding the "Why" Behind the Formula
What Is Atomic Volume?
Atomic volume is not the volume of a single atom — it's the volume occupied by one mole of atoms of an element in its solid state. This is a macroscopic quantity that helps us understand how tightly atoms pack together.
The key formula is:
Atomic Volume=DensityAtomic Mass
Let's break down why this works.
The Core Reasoning: From Mass to Volume
Step 1: What does density tell us?
Density (ρ) is defined as:
ρ=VolumeMass
For a pure solid element, if we take one mole of atoms:
- Mass of one mole = Atomic mass (in g/mol)
- Volume of one mole = Atomic volume (in cm³/mol)
So:
ρ=Atomic volumeAtomic mass
Step 2: Rearranging to find atomic volume
Volume=DensityMass
Therefore:
Atomic Volume=DensityAtomic Mass
Why This Makes Physical Sense
Atomic mass tells you how heavy one mole of atoms is; density tells you how much mass fits in a given space. Dividing mass by density gives the space that mass occupies.
Example intuition: If iron has atomic mass ≈ 56 g/mol and density 7.87 g/cm³, then:
Atomic volume=7.8756≈7.1 cm3/mol
This means one mole of iron atoms (about 6.022×1023 atoms) occupies roughly 7.1 cm³ of space.
Important Exam Points
| Concept | Why It Matters |
|---|---|
| Units | Atomic mass in g/mol, density in g/cm³ → atomic volume in cm³/mol |
| Solid state only | The formula assumes atoms are closely packed; gases/liquids have different packing |
Taking 0.5 A˚ as the atom's radius: r=5×10−11 m.
Vatom=34πr3≈5.24×10−31 m3 …
Treating each hydrogen atom's given 0.5 A˚ size as its radius, one atom occupies about 5.24×10−31 m3; scaled up by Avogadro's number, one mole of hydrogen atoms occupies a total volume of about 3.15×10−7 m3.
Reading the given size correctly
The problem states the size of a hydrogen atom is 0.5 A˚. In this NCERT problem, this value is taken as the atom's radius (the standard atomic-radius quantity used in this kind of estimate), not its diameter:
r=0.5 A˚=0.5×10−10 m=5×10−11 m
Volume of one hydrogen atom (as a sphere)
Vatom=34πr3
r3=(5×10−11)3=125×10−33=1.25×10−31 m3
Vatom=34π×1.25×10−31≈5.24×10−31 m3
Scaling up to one mole
One mole contains NA=6.022×1023 atoms: …
Method: split the calculation into mantissa and power-of-ten separately at every step — cube (or multiply) the plain number and the exponent independently, then recombine. This keeps the error-prone exponent arithmetic clean and gives a built-in order-of-magnitude check on the final answer.
Working exponent and mantissa separately
- Write the radius in mantissa × power-of-ten form. The atom's size, 0.5 A˚, is taken as its radius: r=0.5×10−10 m=5×10−11 m — mantissa 5, exponent 10−11.
- Cube the mantissa and exponent separately. Mantissa: 53=125=1.25×102. Exponent: (10−11)3=10−33. Recombine: r3=1.25×102×10−33=1.25×10−31 m3.
- Multiply by the constant 34π≈4.19 (a pure number, doesn't touch the exponent): Vatom≈4.19×1.25×10−31≈5.24×10−31 m3. …
Common Mistakes: Atomic Volume of a Mole of Hydrogen Atoms
This question gives the hydrogen atom's size as 0.5 Å and, following NCERT's own treatment, this is taken directly as the atom's radius (not its diameter) -- so r=0.5 A˚=5×10−11 m. Getting this one interpretation right or wrong changes the final answer by a factor of 8 (since volume scales as r3), so it's the single most important thing to get right on this problem.
Mistake 1: Treating 0.5 Å as a diameter and halving it again
The error: Reading "size... is about 0.5 Å" as a diameter, then computing radius =0.25 A˚=2.5×10−11 m.
Why it's wrong: For this specific NCERT problem, the given 0.5 Å is the radius -- that is the standard, textbook-prescribed reading. Using it as a diameter shrinks the radius by half and the volume by a factor of 23=8, giving a final answer of roughly 3.9×10−8 m3 instead of the correct ≈3.15×10−7 m3.
How to avoid: Take r=0.5 A˚=5×10−11 m directly, with no extra halving step.
Mistake 2: Forgetting to convert Å to metres before cubing
The error: Plugging r=0.5 (or 5) straight into V=34πr3 without first converting to SI units, since the answer is required in m3.
How to avoid: Always write the conversion explicitly first: 1 A˚=10−10 m, so r=0.5×10−10 m=5×10−11 m, then cube it.
Mistake 3: Forgetting to scale up to one mole
The error: Computing the volume of a single hydrogen atom (≈5.24×10−31 m3) and stopping there -- the question asks for the volume of a mole of atoms. …
- KCET 2024Set D-21 markMCQQ.The ratio of volume of Al27 nucleus to its surface area is (Given R0=1.2×10−15 m ) (A) 2.1×10−15 m (B) 1.3×10−15 m (C) 0.22×10−15 m (D) 1.2×10−15 m
›Reveal solutionSolution
The ratio of volume to surface area for any spherical nucleus is R/3, where R=R0A1/3. For Al27, A=27, so R=1.2×10−15×3=3.6×10−15 m, giving a ratio of 1.2×10−15 m. The correct option is (D).
The key idea here is that a nucleus is modeled as a uniform sphere. For any sphere, the ratio of volume to surface area is simply R/3 — a clean geometric fact. Once you know the nuclear radius formula, the rest is arithmetic.
The nuclear radius is given by R=R0A1/3, where R0=1.2×10−15 m and A is the mass number. For Al27, A=27, so A1/3=3. That gives R=1.2×10−15×3=3.6×10−15 m.
Now, for a sphere:
- Volume V=34πR3
- Surface area S=4πR2
The ratio SV=4πR234πR3=3R.
So the ratio is simply 33.6×10−15=1.2×10−15 m. …
- KCET 2019Set A-11 markMCQQ.In Rutherford experiment, for head-on collision of α-particles with a gold nucleus, the impact parameter is (A) zero (B) of the order of 10−14 m (C) of the order of 10−10 m (D) of the order of 10−6 m
›Reveal solutionSolution
In a head-on collision, the α-particle is aimed directly at the nucleus, so the perpendicular distance between the initial velocity line and the nucleus — the impact parameter — is exactly zero.
The key idea here is the definition of impact parameter in Rutherford's scattering experiment. The impact parameter b is the perpendicular distance between the initial velocity vector of the α-particle and the centre of the target nucleus. It tells you how "off-centre" the collision is.
For a head-on collision, the α-particle is aimed straight at the nucleus. That means the line of the initial velocity passes directly through the centre of the nucleus. The perpendicular distance from that line to the centre is therefore zero.
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Recall the definition: Impact parameter b = distance of closest approach if the nucleus were not there — it's the miss distance. Mathematically, if the initial velocity is along a line, b is the perpendicular distance from the nucleus centre to that line.
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Head-on means zero miss distance: When the α-particle is fired straight at the nucleus, there is no "miss" — it's coming right at it. So b=0.
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Consequences: For b=0, the α-particle experiences the maximum repulsive force (Coulomb force) and comes to rest momentarily at the distance of closest approach before being repelled back. This is the case that gives the largest scattering angle (180∘). …
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