A waste collection tank can hold 18,754 kg of methanol, which has a density of 0.788 g/cm3. What is the volume of the waste collection tank?
Explanation:
Density is defined as the mass of the substance divided by the volume of the substance.
Mathematically, Density = \(\frac{mass}{volume}\)
Since, it is given that mass is 18,754 kg and density is 0.788 g/\(cm^{3}\). We know that there are 1000 grams in 1 kg.
So, 18,754 kg equals 18,754000 grams.
Hence, calculate the volume of the waste collection tank as follows.
Density = \(\frac{mass}{volume}\)
0.788 g/\(cm^{3}\) = \(\frac{18754000 g}{volume}\)
volume = 23799492.38 \(cm^{3}\)
It is known that 1 \(cm^{3}\) equals 0.001 liter. So, 23799492.38 \(cm^{3}\) equals 23799.49 liters.
Thus, we can conclude that volume of the waste collection tank is 23799.49 liters.
Answer : The volume of the waste collection tank is 23799.5 L
Explanation :
Density : It is defined as the mass contained per unit volume.
Formula used for density :
\(Density=\frac{Mass}{Volume}\)
Given :
Mass of methanol = 18754 kg = 18754000 g (1 kg = 1000 g)
Density of methanol = \(0.788g/cm^3\)
Now put all the given values in the above formula, we get:
\(0.788g/cm^3=\frac{18754000g}{Volume}\)
\(Volume=23799492.38cm^3=23799492.38mL=23799.5L\)
Conversion used : \((1cm^3=1mL=0.001L)\)
Therefore, the volume of the waste collection tank is 23799.5 L