Question

What is the true mass of water in vacuum if the apparent mass weighed in air at $\mathbf{24}\mathbf{°}\mathbf{C}$is $\mathbf{1}\mathbf{.}\mathbf{0346}\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{0002}\mathbf{g}$? The density of air is $\mathbf{0}\mathbf{.}\mathbf{0012}\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{0001}\mathbf{g}\mathbf{/}\mathbf{mL}$ and the density of balance weights is $\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{g}\mathbf{/}\mathbf{mL}$. 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

True mass of water in vacuum$\left(\mathrm{m}\right)=1.0357\left(\pm 0.0002\right)\mathrm{g}$

Step-by-step solution

Given

In this task, it is given that the apparent mass weighed in air at $\mathbf{24}\mathbf{°}\mathbf{C}$is $\mathbf{1}\mathbf{.}\mathbf{0346}\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{0002}\mathbf{g}$

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

The formula for buoyancy correction:

$\mathrm{m}=\frac{\mathrm{m}\text{'}\left(1-{\displaystyle \frac{{\mathrm{d}}_{\mathrm{a}}}{{\mathrm{d}}_{\mathrm{w}}}}\right)}{\left(1-{\displaystyle \frac{{\mathrm{d}}_{\mathrm{a}}}{{\mathrm{d}}_{\mathrm{o}}}}\right)}$

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:

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

$$\begin{gathered} \mathrm{m}=1.0357\left(\pm 0.0218\%\right)\mathrm{g} \\ =1.0357\left(\pm 0.0002\right)\mathrm{g} \end{gathered}$$