# A 3.40 kg aluminum ball has an apparent mass of 2.10 kg when submaged in a particular liquid. What is the density of the liquid?

##### 1 Answer

#### Answer:

#### Explanation:

As you know, when an object is *submerged* in a particular liquid, the difference between its mass and its apparent mass will represent the **mass of the liquid** displaced by the object.

You can think of mass in terms of density and *volume*. More specifically, you know that the following relationship exists between mass, volume, and density

#color(blue)("density" = "mass"/"volume")#

This means that the mass of the displaced liquid will be equal to

#rho_"liquid" = m_"liquid"/V_"liquid" implies m_"liquid" = rho_"liquid" * V_"liquid"#

But the **volume** of the displaced liquid will be **equal** to the volume of the ball,which in turn can be written as

#V_"ball" = m_"ball"/rho_"ball"#

Plug this into the above equation to get

#m_"liquid" = rho_"liquid" * m_"ball"/rho_"ball"#

We've already established that the mass of the displaced liquid is the difference between the mass and the apparent mass of the ball, which means that you have

#overbrace(m_"ball" - m_"apparent")^(color(red)(m_"liquid")) = m_"ball" * rho_"liquid"/rho_"ball"#

This means that the density of the liquid will be

#rho_"liquid" = rho_"ball" * (m_"ball" - m_"apparent")/m_"ball"#

Now all that you need is the density of aluminium, which is listed as

#rho_(Al) = "2712 kg/m"^3#

http://www.engineeringtoolbox.com/metal-alloys-densities-d_50.html

Plug in your values and find the density of the liquid

#rho_"liquid" = "2712 kg/m"^3 * ( (3.40 - 2.10) color(red)(cancel(color(black)("kg"))))/(3.40 color(red)(cancel(color(black)("kg"))))#

#rho_"liquid" = color(green)("1040 kg/m"^3)#

The answer is rounded to three sig figs.