# Question 6de52

I found 2.8g/(cm^3#
Density is equal to $\text{mass"/"volume}$ so in our case the volume of the stone will be given observing the increase of volume from 3mL to 14 mL immagining that he only contribution to this increase (of 14-3=11mL) is due to the voume of the stone:
so considering $11 m L = 11 c {m}^{3}$ we have:
Density$= \frac{31}{11} = 2.8 \frac{g}{c {m}^{3}}$