What is the mass of a cube of aluminium \[5.0{\text{cm}}\]on each edge, if the density of aluminium is \[2.7{\text{g/c}}{{\text{m}}^3}\]?
Answer
581.4k+ views
Hint: We should have a knowledge of formula related to volume of the cube. We should also know the relation density, mass and volume.
Volume of a cube is calculated by the formula shown below:
Volume of cube\[ = {{\text{a}}^3}\]
Where a is edge of a cube
The relation of density, mass and volume is shown below:
\[{\text{Density = }}\dfrac{{{\text{Mass}}}}{{{\text{Volume}}}}\]
\[{\text{Mass = Density}} \times {\text{Volume}}\]
Complete answer:
First step is we can calculate the volume of the cube by the use of edge length i.e. a.
For calculation of volume of a cube, we can use a formula as shown below:
Volume of cube\[ = {{\text{a}}^3}\]
Where a is edge of a cube
Now, let’s calculate the volume of a cube:
a\[ = 5.0{\text{cm}}\]
\[{\text{V}} = {{\text{a}}^3}\]
Now we can substitute the known values we get,
\[{\text{V}} = {(5{\text{cm}})^3}\]
On simplification we get,
\[{\text{V}} = 125{\text{c}}{{\text{m}}^3}\]
Now, the second step is to calculate the mass of the aluminium atom. For calculating mass of aluminium atom, we know the relation:
\[{\text{Density = }}\dfrac{{{\text{Mass}}}}{{{\text{Volume}}}}\]
\[{\text{Mass = Density}} \times {\text{Volume}}\]
So,
\[{\text{Density}} = 2.7{\text{ g/c}}{{\text{m}}^3}\]
\[{\text{Volume}} = 125{\text{ c}}{{\text{m}}^3}\]
Now, substitute the values in the above formula,
\[{\text{Mass}} = {\text{Density}} \times {\text{Volume}}\]
\[{\text{m}} = {\text{d}} \times V\]
Now we can substitute the given values we get,
\[{\text{Mass}} = 125{\text{c}}{{\text{m}}^3} \times 2.7{\text{g/c}}{{\text{m}}^3}\]
On simplification we get,
\[{\text{Mass}} = 337.5g\]
Hence, the mass of the aluminium atom is found to be \[337.5g\].
Additional information:
As we know that the aluminium belongs to group number \[13\] of the periodic table. It contains \[3\] valence electrons in its outermost shell. Its mass number is \[27\] and atomic number is \[13.\]
Note:
Always keep in mind, the proper units need to be used. While using the formula for volume of cube, please be careful. It’s better to do the sum by dimensional analysis, so that we always get proper units. Chances of getting wrong units are minimized by the use of dimensional analysis.
Volume of a cube is calculated by the formula shown below:
Volume of cube\[ = {{\text{a}}^3}\]
Where a is edge of a cube
The relation of density, mass and volume is shown below:
\[{\text{Density = }}\dfrac{{{\text{Mass}}}}{{{\text{Volume}}}}\]
\[{\text{Mass = Density}} \times {\text{Volume}}\]
Complete answer:
First step is we can calculate the volume of the cube by the use of edge length i.e. a.
For calculation of volume of a cube, we can use a formula as shown below:
Volume of cube\[ = {{\text{a}}^3}\]
Where a is edge of a cube
Now, let’s calculate the volume of a cube:
a\[ = 5.0{\text{cm}}\]
\[{\text{V}} = {{\text{a}}^3}\]
Now we can substitute the known values we get,
\[{\text{V}} = {(5{\text{cm}})^3}\]
On simplification we get,
\[{\text{V}} = 125{\text{c}}{{\text{m}}^3}\]
Now, the second step is to calculate the mass of the aluminium atom. For calculating mass of aluminium atom, we know the relation:
\[{\text{Density = }}\dfrac{{{\text{Mass}}}}{{{\text{Volume}}}}\]
\[{\text{Mass = Density}} \times {\text{Volume}}\]
So,
\[{\text{Density}} = 2.7{\text{ g/c}}{{\text{m}}^3}\]
\[{\text{Volume}} = 125{\text{ c}}{{\text{m}}^3}\]
Now, substitute the values in the above formula,
\[{\text{Mass}} = {\text{Density}} \times {\text{Volume}}\]
\[{\text{m}} = {\text{d}} \times V\]
Now we can substitute the given values we get,
\[{\text{Mass}} = 125{\text{c}}{{\text{m}}^3} \times 2.7{\text{g/c}}{{\text{m}}^3}\]
On simplification we get,
\[{\text{Mass}} = 337.5g\]
Hence, the mass of the aluminium atom is found to be \[337.5g\].
Additional information:
As we know that the aluminium belongs to group number \[13\] of the periodic table. It contains \[3\] valence electrons in its outermost shell. Its mass number is \[27\] and atomic number is \[13.\]
Note:
Always keep in mind, the proper units need to be used. While using the formula for volume of cube, please be careful. It’s better to do the sum by dimensional analysis, so that we always get proper units. Chances of getting wrong units are minimized by the use of dimensional analysis.
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