Question

A gold atom is heavier than an iron atom;

1 mg

of which element - gold or iron - will contain a greater number of atoms?

Accepted Answer

We know that,

Atomic mass of $Au = 197 u$

Atomic mass of $Fe = 56 u$

Then, gram atomic mass of $Au = 197 g$

Now, $197 g$ of $Au$ contains $= 6.022 \times 10^{23} g$ number of atoms.

Thus, $1 g$ of $Au$ contains $= 6.022 \times 10^{23} /$ ( $197 \times 1$ number of atoms).

$1 g$ of $Au$ contains $= 3.056 \times 10^{21}$ number of atoms.
Multiplying $10^{- 3}$ on both sides
Therefore, $1 mg$ of $Au$ contains $= 3.056 \times 10^{18}$ number of atoms

Then, gram atomic mass of $Fe = 56 g$

Now, $56 g$ of $Fe$ contains $= 6.022 \times 10^{23}$ g number of atoms

Thus, $1 g$ of $Fe$ contains $= 6.022 \times 10^{23} /$ ( $56 \times 1$ number of atoms)

$1 g$ of $Fe$ contains $= 1.0753 \times 10^{22}$ number of atoms
Multiplying $10^{- 3}$ on both sides
$1 mg$ of $Fe$ contains $= 1.0753 \times 10^{19}$ number of atoms
Therefore, $1 mg$ of iron contains more number of atoms than $1 mg$ of gold.

West Bengal Board Solutions for Chapter: Element, Compound and Chemical Reaction, Exercise 7: Exercise 7

West Bengal Board Science Solutions for Exercise - West Bengal Board Solutions for Chapter: Element, Compound and Chemical Reaction, Exercise 7: Exercise 7

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