Question

Calculate the density of a solid with SG $ = 5.70$ in $l{b_m}/f{t^3}$ ? Determine the mass $(kg)$ of $50.0\,L$ of this solid?

Answer 1

Hint:You can solve this question by using a few conversions, like: $1\,l{b_m} = 453.6\,g$ and $1\,ft = 30.48\,cm$ . Some other things that you’ll need to know to solve this question are:
Density of water $\left( {{d_w}} \right) = 1\,g/c{m^3}$ and the formula that:
$SG = \dfrac{{{d_s}}}{{{d_w}}}$
Where, $SG$ is the specific gravity of the substance,
${d_s}$ is the density of the substance and,
${d_w}$ is the density of water

Complete step-by-step solution:
We will try to solve the question with the same approach as told in the hint section of the solution to the given question.
Firstly, let us try to convert the given specific gravity of the solid from $l{b_m}/f{t^3}$ to $g/c{m^3}$ :
For the numerator, we need to convert $5.70\,l{b_m}$ into $g$
We know that $1\,l{b_m} = 453.6\,g$
So, we can see that $5.70\,l{b_m} = 5.70 \times 453.6\,g$
Solving this, we get the numerator as $2585.52\,g$
Now, we will convert the denominator from $f{t^3}$ to $c{m^3}$
We know that $1\,ft = 30.48\,cm$
So, $1\,f{t^3} = 28316.84\,c{m^3}$
Now, the converted specific gravity can be written as: $SG = \,\dfrac{{2585.52}}{{28316.84}}\,g/c{m^3}$
Or, $SG = 0.09\,g/c{m^3}$
By multiplying both numerator and denominator with $1000$ , we get:
$SG = 0.09\,kg/l$
Now, we need to find the density of the solid, which we can find by using the following formula:
${d_s} = \,SG \times {d_w}$
We already found out the value of specific gravity and density of water.
Substituting them in the formula, we get:
${d_s} = 0.09\,kg/l$
Now, let’s find out the mass in the given volume of the solid:
Given volume $V = 50.0\,L$
We know that ${d_s} = \dfrac{{mass}}{{volume}}\,kg/l$
Substituting the values of density and volume of the solid, we get:
$\begin{gathered}
  m = {d_s} \times V \\
  m = 0.09 \times 50 \\
  m = 4.5\,kg \\
\end{gathered} $
Hence, $4.5{\kern 1pt} kg$ is the mass of $50.0L$ of the solid.

Note:- Many students commit mistakes in the conversions while solving the question. Mainly in the part of conversion of the denominator of the specific gravity as we have to convert $f{t^3}$ into $c{m^3}$ and not just $ft$ into $cm$.