Question

How many $lbs$ would a $5.0\,L$ container of mercury weigh$?$ Given: density of mercury is $13.6\,g\,c{m^{ - 3}}$.
(i) $155\,lbs$
(ii) $160\,lbs$
(iii) $150\,lbs$
(iv) $165\,lbs$

Answer 1

Hint:We know that $Density\, = \,\dfrac{{Mass}}{{Volume}}$. The density of mercury and volume of the container are given. Convert one of them to similar units and use the equation to calculate the mass of the mercury container. Convert the mass obtained from the equation to proper unit i.e. $lbs$.

Complete step by step solution:
We need to calculate the mass of the metal container in$lbs$.
We know, $Density\, = \,\dfrac{{Mass}}{{Volume}}........\left( 1 \right)$.
Density of mercury $ = \,\,13.6\,g\,c{m^{ - 3}}$
Volume of the container $ = \,5\,L\, = \,5 \times 1\,d{m^3}\, = \,5 \times {10^3}\,c{m^3}$
Now we can write equation $\left( 1 \right)$ as
$Mass\, = \,Density \times Volume$
Putting the values we get,
$Mass\, = \,13.6\,g\,c{m^{ - 3}} \times 5 \times {10^3}\,c{m^3}\, = \,68000\,g\, = \,68\,kg$
Now, $1\,kg\, = \,2.2046\,lbs$
$\therefore \,Mass\, = \,68\,kg\, \times \,2.2046\,lbs\, = \,149.9\,lbs\, \approx \,150\,lbs$

Hence the correct answer is (iii) $150\,lbs$.

Additional information:Archimedes first used the concept of density to expose the fraud. Density is a physical property of matter that expresses a relationship between mass and volume. If an object contains more mass in a given space, the object is considered to be dense. This relationship is not just about how closely packed together the atoms of an element or the molecules of a compound are, it is also affected by the atomic mass of an element or compound. Since different substances have different densities, density measurements are a useful means for identifying substances. Density is defined as the ratio of an object's mass to its volume. Because it is a ratio, the density of a material remains the same without regard to how much of that material is present. Density is therefore called an intensive property of matter.

Note:The most important step in this question is the conversion of the units. If you make any mistake while converting the units, the calculation will go wrong and hence the result that will be obtained will also be incorrect. Also remember to change the unit of the mass obtained, from $kg$ to $lbs$.