The density of aluminium is 2700kg/m^-3.calculate the mass of a block of aluminium which has a volume of 100cm3?

m^-3

Oct 1, 2017

The mass of the block is 0.270 kg.

Explanation:

There is one error in the problem as written up.
Density = mass/volume, therefore the units should be
either $\frac{k g}{m} ^ 3 \mathmr{and} k g \cdot {m}^{-} 3$.

This will require converting the volume units from $c {m}^{3} \text{ to } {m}^{3.}$ I will use a method I call "multiplying by 1" that I have used since my first Physics class. You may go to this site for an explanation of my method:
https://socratic.org/questions/a-mile-is-5280-ft-long-1-ft-is-approximately-0-305-m-how-many-meters-are-there-i#469538

Converting $100 c {m}^{3}$ step by step:
$100 c {m}^{3} \cdot {\left(\frac{1 m}{100 c m}\right)}^{3}$
$100 c {m}^{3} \cdot {\left(1 m\right)}^{3} / {\left({10}^{2} c m\right)}^{3}$
$1 \cdot {10}^{2} c {m}^{3} \cdot \frac{1 {m}^{3}}{{10}^{6} c {m}^{3}}$
$1 \cdot \cancel{{10}^{2}} \cancel{c {m}^{3}} \cdot \frac{1 {m}^{3}}{{10}^{4} \cdot \cancel{{10}^{2}} \cancel{c {m}^{3}}} = 1 \cdot {10}^{-} 4 {m}^{3}$

Density = mass/volume, therefore one algebraic operation will tell us that
mass = density*volume. So, the last step to finding your answer is

$\text{mass} = \frac{2700 k g}{\cancel{{m}^{3}}} \cdot 1 \cdot {10}^{-} 4 \cancel{{m}^{3}} = 0.270 k g$

I hope this helps,
Steve