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 ${24.0}^{o} C$ 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.

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.

${\rho}_{{H}_{2} O} = \left(0.997296 \text{g")/("mL}\right)$

$85.2236 - 60.1324 = 25.0192 \text{g}$

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

${V}_{{H}_{2} O} = 25.0192 \text{g" * (0.997296"g")/("mL") approx 25.0234"mL}$

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.