A stone is placed in the bottom of an empty graduated cylinder. When 35.0 mL of water is added the level of the water reaches the 50.0 mL mark. If the density of the stone is 3.5 g/#cm^3#, what is the volume of the stone, and what is its mass?

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Answer 1 · 2016-09-19T08:27:27

Clearly the volume of the stone is #15*mL#.

The volume of the stone is #15*mL#, because when #35*mL# water were added, the level rose to the #50*mL# mark.

Now since #"density, "rho# #=# #"Mass"/"Volume"#,

#"Mass"# #=# #rhoxx"volume"# #=# #3.5*g*cm^-3xx15*mLxx1*cm^3*mL^-1=52.5*g#.

Here we used the relationship that #1*mL# #=# #1*cm^-3#. How?

Well #1*cm^-3# #=# #1*(10^-2*m)^3# #=# #1xx10^-6*m^3# #=# #10^-3xx10^-3*m^3# #=# #10^-3*L# #=# #1*mL#, because there are #1000*L# in a #m^3# volume.