Question

An unknown and unreactive solid has a density of 0.563 g/mL. It is placed in a glass of gasoline. How much of this solid will be above the liquid gasoline if the specific gravity of gasoline is 0.66?

Answer 1

$$\rho_{solid}=563\frac{kg}{m^3},\rho_{gas}=660\frac{kg}{m^3}$$

$$\rho_{solid}gV=\rho_{gas}gV'$$

$$\frac{V-V'}{V}=1-\frac{\rho_{solid}}{\rho_{gas}}$$

$$\frac{V-V'}{V}=1-\frac{563}{660}=0.147\text{ or 14.7\%}$$