Question

A metal M of atomic weight 60 gm/mol has a density of 10.0 g/cc. Calculate the volume occupied by one atom.
(A)${{1 \times 1}}{{\text{0}}^{{\text{ - 23}}}}{\kern 1pt} {\text{c}}{{\text{m}}^{\text{3}}}$
(B)${{\{2 \times 1}}{{\text{0}}^{{\text{ - 23}}}}{\kern 1pt} {\text{c}}{{\text{m}}^{\text{3}}}$
(C)${{3 \times 1}}{{\text{0}}^{{\text{ - 23}}}}{\kern 1pt} {\text{c}}{{\text{m}}^{\text{3}}}$
(D)${{4 \times 1}}{{\text{0}}^{{\text{ - 23}}}}{\kern 1pt} {\text{c}}{{\text{m}}^{\text{3}}}$

Answer 1

Hint: Any metallic substance which weighs its atomic weight or a compound of one mole weighing its molar mass consists of Avogadro number of atoms or molecules of the substance. It is obtained by division of charge on a mole of electrons by charge on a single electron.

Complete Step-by-step solution:
Avogadro number is a standard value of number of particles (atoms or molecules) per mole which is ${\text{6}}{{.022 \times 1}}{{\text{0}}^{{\text{23}}}}{\kern 1pt} {\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$.
Given data is: Atomic weight of the metal is 60 gm / mol,
The density of the metal is 10 g /cc. We know that the density of a substance is mass per unit volume. But the given mass is of one mole of the substance. So let us find the volume of metal per a mole of it from the formula of density, $\rho = \dfrac{m}{V}$.
Substituting the value of mass and volume in the above formula, $\begin{gathered}
{\text{10 = }}\dfrac{{{\text{60}}}}{{\text{V}}} \\
\end{gathered} $
 V= 6ml or 6cc.
From the volume of 6ml let us find out the volume for one mole of the metal by criss-cross method. If Avogadro number of atoms occupies 6ml of the volume then one atom occupies-\[\begin{gathered}
{{\text{V}}_{\text{1}}}{\text{ = }}\dfrac{{\text{6}}}{{{\text{6}}{{.022 \times 1}}{{\text{0}}^{{\text{23}}}}}} \\
{{\text{V}}_{\text{1}}}{{ = 1 \times 1}}{{\text{0}}^{{\text{ - 23}}}}{\kern 1pt}
{\text{c}}{{\text{m}}^{\text{3}}} \\
\end{gathered} \]

From the above calculation the answer for volume occupied by one atom of the given metal is option (A).

Note: A substance which contains Avogadro number of particles is termed as one mole of it and it weighs its atomic weight, occupies 22.4lt at NTP and 22.4lt at STP. If the substance is a compound of more than one element then it weighs its molecular mass or molar mass.