Chapter 2, Problem 2.26P

(a)

Interpretation Introduction

Interpretation:

The true mass water in vacuum should be calculated.

Concept introduction:

Buoyancy Correction:

The weight of the object in air is less then weight of the object in vacuum because of buoyancy, this weight variation is corrected using Buoyancy equation it is,

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

Where,

$\text{m}$ is mass of object in vacuum.

$\text{m'}$ is mass of object in air (read on a balance).

${\text{d}}_{\text{a}}$ is the density of air $\left(\text{0}\text{.001 2 g/mL near I bar and 25°C}\right)$.

${\text{d}}_{\text{w}}$ is the density of calibration weights $\text{(8}\text{.0 g/mL)}$.

$\text{d}$ is the density of the object being weighed.

Answer to Problem 2.26P

True mass the mass of $\text{50}\text{.037 mL}$ water at $\text{20}°\text{C}$ in vacuum is $\text{50}\text{.0026}\text{g}$

Explanation of Solution

To calculate the mass of $\text{50}\text{.037 mL}$ water at $\text{20}°\text{C}$

Given,

Volume of water at $\text{20}°\text{C}$ in $\text{50}\text{mL}$ flask is $\text{50}\text{.037 mL}$

Density of water $\text{20}°\text{C}$ is $\text{0}\text{.998 2071 g/mL}$

Mass of $\text{50}\text{.037 mL}$ water at $\text{20}°\text{C}$ is, $\text{49}\text{.95}\text{g}$

$\begin{array}{l}\text{Mass}\text{=}\text{Density×volume}\\ \text{=}\text{0}\text{.9982071 g/mL×}\text{50}\text{.037}\text{mL}\\ \text{=}\text{49}\text{.95}\text{g}\end{array}$

To calculate true mass of water in vacuum at $\text{20}°\text{C}$

Mass of $\text{50}\text{.037 mL}$ water at $\text{20}°\text{C}$ is, $\text{49}\text{.95}\text{g}$

The density of air is $\text{0}\text{.0012 g/mL}$

Density of balance weight is $\text{8}\text{.0 g/mL}$

We know,

$\begin{array}{l}\text{Mass}\text{of}\text{water}\text{in}\text{vacuum}\text{=}\frac{\text{(49}\text{.95}\text{g}\text{)}\left(\text{1-}\frac{\text{0}\text{.0012}\text{g/}\text{mL}}{\text{8}\text{.0}\text{g/mL}}\right)}{\left(\text{1-}\frac{\text{0}\text{.0012}\text{g/mL}}{\text{0}\text{.9982071}\text{g/mL}}\right)}\\ \text{=}\text{50}\text{.0026}\text{g}\end{array}$

The calculated mass and density of water, density of air, and density of balance weight are plugged in above equation to give true mass of water in vacuum.

True mass of water in vacuum at $\text{20}°\text{C}$ is $\text{50}\text{.0026}\text{g}$

Conclusion

The true mass water in vacuum should be calculated.

(b)

Interpretation Introduction

Interpretation:

The true mass water at air should be calculated.

Concept introduction:

Buoyancy Correction:

The weight of the object in air is less then weight of the object in vacuum because of buoyancy, this weight variation is corrected using Buoyancy equation it is,

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

Where,

$\text{m}$ is mass of object in vacuum.

$\text{m'}$ is mass of object in air (read on a balance).

${\text{d}}_{\text{a}}$ is the density of air $\left(\text{0}\text{.001 2 g/mL near I bar and 25°C}\right)$.

${\text{d}}_{\text{w}}$ is the density of calibration weights $\text{(8}\text{.0 g/mL)}$.

$\text{d}$ is the density of the object being weighed.

Answer to Problem 2.26P

The true mass water in air be calculated. $\text{0}\text{.2963}\text{g}$

Explanation of Solution

To calculate the true mass water at air.

Given,

Volume of water at $\text{20}°\text{C}$ in $\text{50}\text{mL}$ flask is $\text{50}\text{.037 mL}$

Density of water $\text{20}°\text{C}$ is $\text{0}\text{.998 2071 g/mL}$

Mass of $\text{50}\text{.037 mL}$ water at $\text{20}°\text{C}$ is, $\text{49}\text{.95}\text{g}$

$\begin{array}{l}\text{Mass}\text{=}\text{Density×volume}\\ \text{=}\text{0}\text{.9982071 g/mL×}\text{50}\text{.037}\text{mL}\\ \text{=}\text{49}\text{.95}\text{g}\end{array}$

The true mass of water in air at $\text{20}°\text{C}$ is $\text{49}\text{.95}\text{g}$

Conclusion

The true mass of water in air at $\text{20}°\text{C}$ was determined.