Question

a researcher dispenses distilled deionized water from a 20 - ml transfer pipet into an empty 8.4376 - g weighting bottle. if the total mass of water and weighting bottle is 28.5845 g, what is the volume of the water delivered by the 20 - ml pipet? the density of water is 0.9967867 g/ml. given that the density of air is expected to be 0.0011917 g/ml and that of the calibration weights is expected to be 7.95 g/ml at the lab temperature near 23.3c temperature, answer the following a. the buoyancy correction factor, b, is expected to be 1.0010 (report to 4 decimal places). it is unitless. b. the sample mass, m, is 20.1469 g. report to the expected number of decimal places. units are known to be g, so you neednt report the units. c. correct the mass for buoyancy. report the corrected mass, m g. show one extra significant digit d. calculate the volume delivered by the pipette to 4 significant figures.

Explanation:

Step1: Identify the formula for mass - buoyancy correction

The formula for correcting mass for buoyancy is $m = m'\times B$, where $m$ is the corrected mass, $m'$ is the sample mass and $B$ is the buoyancy correction factor.

Step2: Substitute the given values

We are given that $m'=20.1469$ g and $B = 1.0010$. So, $m=20.1469\times1.0010$.
$m = 20.1469\times1.0010=20.1670469$ g. Rounding with one extra significant digit, $m = 20.167$ g.

Step3: Use the density - mass - volume formula

The formula for density $
ho=\frac{m}{V}$, where $
ho$ is density, $m$ is mass and $V$ is volume. We can re - arrange it to $V=\frac{m}{
ho}$. The density of water $
ho = 0.9967867$ g/mL and the corrected mass $m = 20.167$ g.

Step4: Calculate the volume

$V=\frac{20.167}{0.9967867}\approx20.23$ mL.

Answer:

c. 20.167
d. 20.23