Question

1.5 slug plastic leads are poured in a tank containing 8000 gal oil. Determine the concentration of plastic leads in the oil in mg/L. The specific gravities of the plastic leads and the oil are, respectively, 1.2 and 0.8. The volume of plastic leads is (a) neglected, (b) not neglected

Answer 1

Solution :-

8000 gal oil

1.5 slug

Specific gravity oil = 0.8 and slug = 1.2

Lets first calculate the mass of the slug

1.5 slug * 14594 g / 1 slug = 21891 g

21891 g * 1000 mg / 1 g = 2.1891*10^7 mg slug

Now lets calculate the volume of the slug

Volume = mass /density

              = 21891 g / 1.2 g /ml

             = 18243 ml

18243 ml * 1 L / 1000 ml = 18.243 L

Now lets find volume of oil in liter

8000 gal * 3.78541 L/ 1 gal = 30283.28 L

Total volume = 18.243 L + 30283.28 L = 30301 .52 L

Now lets calculate the concentration in mg/L

2.1891 *10^7 mg plastic / 30301.52 L = 722.45 mg/ L

So the concentration is 722.45 mg / L

The volume of the slug cannot be neglected