Question

A metal M of atomic weight 60gm /mol has density of 10.0 g/cm^3calculate the volume occupied by one atom
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Answer 1

first find volume occupied by one mole of atom it will be 60/10=6cm^3

now one mole contain 6.022×10^23 atoms

so volume of one atom is

6/(6.022×10^23)

10^-23cm^3(approx)

Answer 2

Answer:

The volume occupied by one atom of metal is equal to  10⁻²³cm³.

Explanation:

Given, the atomic mass of the metal = 60g

The density of the metal = 10.0 g/cm³

As we know density of one mole metal will be equal to ratio of atomic mass of metal to volume occupied by one mole metal atoms.

We can calculate the volume from the formula of density:

Density = Mass/ volume

The volume of the metal = mass/ density

The volume of the metal $=\frac{60g/mol}{10gcm^{-3}} = 6cm^3/mol$

Therefore, the volume occupied by 1 mole of metal = 6cm³

The volume of one atom of metal $=\frac{6}{6.02\times 10^{23}} = 1\times 10^{-23}cm^3$

Therefore, the volume of one atom of metal is equal to 10⁻²³cm³.

To know more about "S.I. unit of density"

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To know more about "Mole concept"

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