Given:
Mass of methanol, m = 18754 kg
Density of methanol, ρ = 0.788 g/cm³
By definition, the volume of methanol in the collection tank is
Volume = mass/density
[tex]= \frac{(18754 \, kg)*(10^{3} \, \frac{g}{kg}) }{(0.788 \, \frac{g}{cm^{3})} } \\ \\ =2.38 \times 10^{7} \, \frac{g}{cm^{3}} [/tex]
Answer: 2.38 x 10⁷ g/cm³
