Chapter 2, Problem 2.83QA

Interpretation Introduction

To find:

To compare the number of atoms of uranium and diamond in a cube (having 1 cm on a side) made of each of these two elements.

Answer to Problem 2.83QA

Solution:

The cube made up of diamond contains more atoms.

Explanation of Solution

Concept:

The volume of two cubes made up of Uranium and Carbon are same. Only these two elements have different values of densities.

So, the masses of uranium and diamond cubes will be different since

Mass = Volume x Density

So, here, firstly, the given masses of each cube have to be calculated separately by using the above formula.

Atomic masses of Uranium and Carbon are known.

A mole is the SI unit of amount chemical substance. When writing units, it is written as “mol”.

Formula:

i) $\mathrm{N}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{a}\mathrm{n}\mathrm{y}\mathrm{ }\mathrm{s}\mathrm{u}\mathrm{b}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{ }=\mathrm{ }\frac{\mathrm{g}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{ }\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{ }\mathrm{s}\mathrm{u}\mathrm{b}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}}{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{ }\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{ }\mathrm{s}\mathrm{u}\mathrm{b}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}}$

ii) The number of atoms of each element = Number of moles x Avogadro number

Given information:

Density of Uranium = $19.05$ g/cm³

Density of Carbon (Diamond) = $3.514$ g/cm³

Length of each side of cube = $1$ cm

Atomic mass of Uranium (U) = $238.03$ g/mol

Atomic mass of Diamond = $12.01$ g/mol

Calculation:

We will calculate the volume of cube from the 1 cm edge length.

$\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{c}\mathrm{u}\mathrm{b}\mathrm{e}\mathrm{ }=\mathrm{ }1\mathrm{ }\mathrm{c}\mathrm{m}\mathrm{ }\times \mathrm{ }1\mathrm{ }\mathrm{c}\mathrm{m}\mathrm{ }\times \mathrm{ }1\mathrm{ }\mathrm{c}\mathrm{m}\mathrm{ }=\mathrm{ }1\mathrm{ }{\mathrm{c}\mathrm{m}}^3$

We will calculate the mass of uranium from volume and density.

$1\mathrm{ }{\mathrm{c}\mathrm{m}}^3\mathrm{ }\times \mathrm{ }\frac{19.05\mathrm{ }\mathrm{g}}{{\mathrm{c}\mathrm{m}}^3}=19.05\mathrm{ }\mathrm{g}\mathrm{ }\mathrm{U}\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{i}\mathrm{u}\mathrm{m}$

Now calculate the moles of uranium atom by using $19.05$ g and atomic weight.

$19.05\mathrm{ }\mathrm{g}\mathrm{ }\mathrm{U}\mathrm{ }\times \frac{1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}}{238.03\mathrm{ }\mathrm{g}\mathrm{ }\mathrm{U}}\mathrm{ }=\mathrm{ }0.08003\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{ }\mathrm{U}$

We will calculate the number of atom by using the Avogadro’s number.

$0.08003\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }\mathrm{U}\mathrm{ }\times \mathrm{ }\frac{6.022\mathrm{ }\times \mathrm{ }{10}^{23}\mathrm{ }\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{s}}{1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}}=4.\mathrm{ }819\mathrm{ }\times \mathrm{ }{10}^{22}\mathrm{ }\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{s}$.

So $4.\mathrm{ }819\mathrm{ }\times \mathrm{ }{10}^{22}\mathrm{ }\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{s}$ present in uranium atom

Now we will calculate the mass of diamond from volume and density.

$1\mathrm{ }{\mathrm{c}\mathrm{m}}^3\mathrm{ }\times \mathrm{ }\frac{3.514\mathrm{ }\mathrm{g}}{{\mathrm{c}\mathrm{m}}^3}=3.514\mathrm{ }\mathrm{g}\mathrm{ }\mathrm{D}\mathrm{i}\mathrm{a}\mathrm{m}\mathrm{o}\mathrm{n}\mathrm{d}$

Now calculate the moles of uranium atom by using 3.514 g and atomic weight.

$3.514\mathrm{ }\mathrm{g}\mathrm{ }\mathrm{D}\mathrm{i}\mathrm{a}\mathrm{m}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{ }\times \frac{1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}}{12.01\mathrm{ }\mathrm{g}}\mathrm{ }=\mathrm{ }0.2926\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{ }\mathrm{D}\mathrm{i}\mathrm{a}\mathrm{m}\mathrm{o}\mathrm{n}\mathrm{d}$

We will calculate the number of atom by using the Avogadro’s number.

$0.2926\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }\mathrm{D}\mathrm{i}\mathrm{a}\mathrm{m}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{ }\times \mathrm{ }\frac{6.022\mathrm{ }\times \mathrm{ }{10}^{23}\mathrm{ }\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{s}}{1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}}=1.762\mathrm{ }\times \mathrm{ }{10}^{23}\mathrm{ }\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{s}$

So, the cube of diamond contains more number of atoms than uranium.

Conclusion:

We calculate the number of atom from the edge length of cube, atomic mass and Avogadro’s number.