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Question: A graduated cylinder has a mass of \( 50 \cdot g \) when empty. When \( 30 \cdot mL \) of water is a...

A graduated cylinder has a mass of 50g50 \cdot g when empty. When 30mL30 \cdot mL of water is added, it has a mass of 120g120 \cdot g . If a rock is added to the graduated cylinder, the level rises to 75ml75 \cdot ml and the total mass is now 250g250 \cdot g . What is the density of the rock?

Explanation

Solution

Density is the ratio of mass and volume of the substance. It is the measure of how densely the particles of a substance are packed. For the given problem, mass and volume of the rock is needed to be found and then mass has to be divided by the volume to obtain density.

Complete answer:
The given mass of the system before the addition of rock is 120g120 \cdot g and the mass after the addition of 250g250 \cdot g . So, the final mass of rock can be calculated as the difference between both the values.
\eqalign{ & mas{s_{rock}} = mas{s_{after}} - mas{s_{before}} \cr & mas{s_{rock}} = 250 \cdot g - 120 \cdot g \cr & mas{s_{rock}} = 130 \cdot g \cr}
Similarly, the initial volume of the system before the addition of rock is given as 30mL30 \cdot mL and the final volume of the system after the addition of rock is given as 75ml75 \cdot ml . So, the volume of rock can be calculated as the difference between both the values.
\eqalign{ & volum{e_{rock}} = volum{e_{after}} - volum{e_{before}} \cr & volum{e_{rock}} = 75 \cdot ml - 30 \cdot ml \cr & volum{e_{rock}} = 45 \cdot ml \cr}
Now, the density of rock can be calculated as the ratio of mass and volume,
\eqalign{ & density = \dfrac{{mass}}{{volume}} \cr & density = \dfrac{{130 \cdot g}}{{45 \cdot ml}} \cr & density = 2.9 \cdot g/ml \cr & density = 2.9 \cdot g/c{m^3} \cr}
Hence, the density of rock is calculated as 2.9g/cm32.9 \cdot g/c{m^3} .

Note:
The density is expressed in the unit of gram per millilitre or gram per centimeter cube. Centimeter cube is equivalent to a millilitre. The first given mass of the cylinder has no direct role in this calculation, so don’t get confused.