Supplemental Instruction-Chem 120: Dimensional Analysis Using Density

Uploaded by UTKSSC on 18.01.2012

Music. Hi I'm Elizabeth and I'm a Chem 120 SI Leader. Today we're going over dimensional
analysis using density. Today we're doing a dimensional analysis problem involving density.
Given the volume and mass of an unknown liquid the volume is 25.0 mL, the mass is 0.02195
Kg. We're going to use these variables to find the density. We're also given the density
of Benzene, which is 878.7 grams per liter. After we find the density for the unknown,
we'll see if it matches the density of Benzene to discover what our unknown liquid is. First
off we're going to change kilograms into grams. We know that one kilogram is a thousand grams.
You put the kilograms on the bottom to mark them out. Now you want to change this density
of Benzene into grams per milliliters. You put liters on the top to mark it out, and
one liter is...(laughs) a thousand milliliters. This gives us 0.8787 grams per milliliter.
Now we're going to see if the unknown liquid matches up with the density of Benzene. Oops
we don't need a zero there, mark that out. 21.95 grams over 25.0 milliliters equals 0.878
grams per milliliter. That matches up so our unknown liquid is Benzene. This has been a
dimensional analysis problem. Thank you!