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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?

Verified Answer :

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