# A certain sold has a density of 8.0 "g/cm"^3. If 16.0 g of this solid are put into 4.00 mL of water, which drawing below most closely represents the volume of water after the solid is added?

Mar 29, 2015

The idea behind placing a solid in a known volume of water is that the volume of water it will displace will be proportional to its mass and density. Density is defined as mass per unit of volume. SInce you know the solid's density and its mass, you can easily solve for its volume by

rho = m/V => V = m/(rho) = (16.0cancel("g"))/(8.0cancel("g")/"cm"^3) = "2.0 cm"^3

If you take into account the fact that ${\text{1 cm}}^{3}$ is equal to $\text{1 mL}$. the volume of the solid will be

2.0cancel("cm"^3) * ("1 mL")/(cancel("1 cm"^3)) = "2.0 mL"

Since the water has a volume of 4.0 mL, adding the solid will raise the water level in the beaker to

V_("total") = "2.0 mL" + "4.0 mL" = "6.0 mL"

Therefore, pick a drawing that shows that the total volume in the beaker increased by 50% - from 4.0 mL to 6.0 mL.