Chapter 3: Q19P (page 62)

What is the true mass of water in vacuum if the apparent mass weighed in air at $24 ° C$ is $1.0346 \pm 0.0002 g$ ? The density of air is $0.0012 \pm 0.0001 g / mL$ and the density of balance weights is $8.0 \pm 0.5 g / mL$ . The uncertainty in the density of water in Table 2-7 is negligible in comparison to the uncertainty in the density of air.

Short Answer

Expert verified

True mass of water in vacuum $(m) = 1.0357 (\pm 0.0002) g$

Step by step solution

Given

In this task, it is given that the apparent mass weighed in air at $24 ° C$ is $1.0346 \pm 0.0002 g$

Using formula for buoyancy correction find value of mass of water in vacuum

The formula for buoyancy correction:

$m = \frac{m' (1 - \frac{d_{a}}{d_{w}})}{(1 - \frac{d_{a}}{d_{o}})}$

Here $d_{a}$ stands for the density of air, whereas $d_{o}$ stands for the density of weighed product in air.

Using the values given in the task the following true mass would be:

$m = 1.0346 (\pm 0.0002) g \times \frac{(1 - 0.0012 (\pm 0.0001) / 8 (\pm 0.05))}{(1 - (0.0012 (\pm 0.0001) / 0.9972995))} m = 1.0346 (\pm 0.0002) g \times \frac{(1 - 0.0012 (\pm 0.0001) / 8 (\pm 0.05))}{(1 - (0.0012 (\pm 0.0001) / 0.9972995))} m = 1.0346 (\pm 0.0019 %) g \times \frac{(1 - 0.0012 (\pm 8.33 %) / 8 (\pm 6.25 %))}{(1 - (0.0012 (\pm 8.33 %) / 0.9972995))} m = 1.0346 (\pm 0.0019 %) g \times \frac{1 - 0.000150 (\pm 0.0000156)}{1 - 0.001203 (\pm 000100} m = 1.0346 (\pm 0.0019 %) g \times \frac{0.9998500 (\pm 0.00156 %)}{0.998797 (\pm 0.0100 %}$