Given the following, what the actual volume of the pipet?

You are asked calibrate a 25.00-mL volumetric pipet. You determine the temperature of your distilled water is exactly #24.0^oC# You carefully determined the mass of a clean dry beaker and found it was 60.1324 g. You pulled water up to the mark and transferred this to the beaker and found the new mass was 85.2236 g.

1 Answer
Sep 4, 2017

Somewhere in your readings or notes should be the density of water at this temperature you reference. I found it in a table as these values are usually tabulated along with other important data for chemistry calculations.

#rho_(H_2O) = (0.997296"g")/("mL")#

#85.2236 - 60.1324 = 25.0192"g"#

This difference in mass is the mass of water you pipetted. Let's convert it to volume with the density:

#V_(H_2O) = 25.0192"g" * (0.997296"g")/("mL") approx 25.0234"mL"#

This deviation from the stated volume of the pipet is likely user error, but for the purposes of this question, the "actual" volume is thus.