Question

A metal M of atomic weight 54.94 has a density of 7.42 g/cc .calculate the apparent volume occupied by one atom of metal. ​

Answer 1

Answer : The apparent volume occupied by one atom of metal is, $1.23\times 10^{-23}cm^3$

Explanation : Given,

Atomic weight of metal = 54.94 g/mole

Density of metal = $7.42g/cm^3$

First we have to calculate the mass of 1 atom of metal.

$\text{Mass of 1 atom of metal}=\frac{\text{Atomic weight of metal}}{\text{Avagadro's Number}}$

Now put all the given values in this formula, we get the mass of 1 atom of metal.

$\text{Mass of 1 atom of metal}=\frac{54.94g/mole}{6.022\times 10^{23}mole}=9.12\times 10^{-23}g$

Now we have to calculate the apparent volume occupied by one atom of metal.

Formula used :

$Density=\frac{Mass}{Volume}$

Now put all the values in this formula, we get the apparent volume occupied by one atom of metal. ​

$7.42g/cm^3=\frac{9.12\times 10^{-23}g}{Volume}$

$Volume=1.23\times 10^{-23}cm^3$

Therefore, the apparent volume occupied by one atom of metal is, $1.23\times 10^{-23}cm^3$

Answer 2

Answer:

The Answer should be 1.23 × 10‐²³ cm³