A graduated cylinder if filled with water to a level of 40.0 mL. when a piece of paper of copper is lowered into the cylinder, the water level rises to 63.4mL. find the volume of the copper sample. if the density of the copper is 8.9g/cm³, what is the mass

The level of the water without the copper is 40.0 mL.

After placing the copper, the water reached to 63.4 mL.

So the displacement is

Displacement = 63.4 - 40.0

Displacement = 23.4 mL

The displacement is equal to the volume of the copper.

The volume of the copper is 23.4 cm^3.

The formula of the density is

Density = Mass / Volume

8.9 g/cm^3 = Mass / 23.4 cm^3

Mass = 8.9 / 23.4

Mass = 0.38034 g

Mass = 3.8 x 10^-1 g

So the mass is 3.8 x 10^-1 g and the volume is 23.4 cm^3

Given:

Water level is 40mL

Water level rises to 63.4mL after copper is placed

Density is 8.9g/cm3

Required:

Mass of the object

Solution:

The displaced water level is the volume of the copper

63.4mL - 40mL = 23.4mL

Since 1mL = 1cm3, therefore 23.4mL is 23.4cm3

The density formula is D = M/V where D is density, M is mass of the object and V is the volume of the object. Rearranging we get M = DV.

M = DV

M = (8.9g/cm3) (23.4cm3)

M = 208 g